this is just a sample for my activity in school

1. Acids, Bases and Salts
All substances are acidic, neutral or basic (alkaline). How acidic or basic a substance is shown by its pH. There are
several other ways by which we could find out whether a substance is acidic, neutral or basic.
pH Scale:
This is a scale that runs from 0 to 14. Substances with a pH below 7 are acidic. Substances with pH above 7 are
basic. And those with pH 7 are neutral.
Indicators:
Indicators are substances that identify acidity or alkalinity of substances. They cannot be used in solid form.
Universal Indicator:
This is a substance that changes color when added to another substance depending on its pH. The indicator and
the substance should be in aqueous form.
Litmus Paper or Solution:
This indicator is present in two colors: red and blue. We use blue litmus if we want test a substance for acidity. We
use red litmus if we want to test a substance for alkalinity. Its results are:
 Acids: Turns blue litmus paper/ solution red,
 Bases: Turns red litmus paper/ solution blue,
 Neutral: if it is used as paper the color doesn’t change. If it is used as solution it turns purple.
Note: use damp litmus paper if testing gases.
Phenolphthalein:
This is an indicator that is used to test for alkalinity because it is colorless if used with an acidic or neutral substance
and it is pink if it is used with a basic substance.
Methyl Orange:
This indicator gives fire colors: Red with acids, yellow with neutrals and orange with bases.
Acids:
Acids are substances made of a hydrogen ion and non-metal ions. They have the following properties:
 They dissolve in water producing a hydrogen ion H+,
 They have a sour taste,
 Strong ones are corrosive,
 Their pH is less than 7.
All acids must be in aqueous form to be called an acid. For example Hydrochloric acid is hydrogen chloride gas
dissolved in water. The most common acids are:
 Hydrochloric acid HCl,
 Sulphuric Acid H2SO4,
 Nitric Acid HNO3,
 Cirtric Acid,
 Carbonic Acid H2CO3.
Strength of Acids:

2. One of the most important properties of acids is that it gives hydrogen ion H+ when dissolved in water. This is why
the amount of H+ ions the acid can give when dissolved in water is what determines its strength. This is
calledionization or dissociation. The more ionized the acid is the stronger it is, the lower its pH. The more H+ ions
given when the acid is dissolved in water the more ionized the acid is.
Strong Acids:
they give large amounts of
H+ ionsExamples:
Acid
Weak Acids:
they give small amounts of
H+ ionsExamples:
(CH3COOH)
Hydrochloric acid is a strong acid. When it is dissolved in water all HCl molecules are ionized into H+ and Cl-
ions. It is fully ionized.
Ethanoic acid has the formula CH3COOH. It is a weak acid. When it is dissolved in water, only some of the
CH3COOH molecules are ionized into CH3COO- and H+ ions. It is partially ionized.
Note: Acids with pH 3 or 4 can be considered moderate in strength.
Solutions of strong acids are better conductors of electricity than solutions of weak acids. This is because they
contain much more free mobile ions to carry the charge.
Concentrated acids are not necessarily strong. The concentration of an acid only means the amount of molecules of
the acid dissolved in water. Concentrated acids have a large amount of acid molecules dissolved in water. Dilute
acids have a small amount of acid molecules dissolved in water. Concentration is not related to strength of the
acids. Strong acids are still strong even if they are diluted. And weak acids are still weak even if they are
concentrated.
Bases:
Bases are substances made of hydroxide OH- ions and a metal. Bases can be made of:
 Metal hydroxide (metal ion & OH- ion)
 Metal oxides
 Metal carbonates (metal ion & CO32-)
 Metal hydrogen carbonate (Bicarbonate)
 Ammonium hydroxide (NH4OH)
 Ammonium Carbonate ((NH4)2CO3)
Properties of bases:
 Bitter taste
 Soapy feel
 Have pH’s above 7
 Strong ones are corrosive
Some bases are water soluble and some bases are water insoluble. Water soluble bases are also called alkalis.
Like acids, alkalis' strength is determined by its ability to be ionized into metal and hydroxide OH- ions. Completely
ionized alkalis are the strongest and partially ionized alkalis are the weakest. Ammonium hydroxide is one of the
strongest alkalis while weak alkalis include the hydroxides of sodium, potassium and magnesium.
Types of Oxides:
Basic Oxides
oxides
forming a salt and water
Amphoteric
Oxides
Aluminum, Zinc & Lead
Acidic Oxides
-
metal oxides except non-
metal monoxides

3. water except group 1
metal oxides.
acid forming salt and
water
2O,
CaO and CuO
when reacting with an
alkali & vice versa
hydroxides are
amphoteric too
water when reacting with
an acid or an alkali.
alkali to form salt and
water
monoxides are neutral
oxides
2, NO2,
SO2 (acidic oxides) &
CO, NO,
H2O (neutral oxides)
Salts:
A salt is a neutral ionic compound. Salts are one of the products of a reaction between an acid and a base. Salts are
formed in reactions I n which the H+ ion from the acid is replaced by any other metal ion. Some salts are soluble in
water and some are insoluble.
Soluble Salts:
and PbCl2
CaSO4, BASO4, PbSO4
Insoluble Salts:
lead chlorides
(AgCl & PbCl2)
sulphates (CaSO4, BASO4,
PbSO4)
group 1 metals and
ammonium carbonates
Preparing Soluble Salts:
Displacement Method (Excess Metal Method):
Metal + Acid → Salt + Hydrogen
Note: this type of method is suitable to for making salts of moderately reactive metals because highly reactive
metals like K, Na and Ca will cause an explosion. This method is used with the MAZIT (Magnesium, Aluminum,
Zinc, Iron and Tin) metals only.
Example: set up an experiment to obtain magnesium chloride salt.
Mg + 2HCl → MgCl2 + H2
1. Add 100 cm3 of dilute hydrochloric acid to a beaker
2. Add excess mass of powdered magnesium
3. When the reaction is done, filter the mixture to get rid of excess magnesium (residue)
4. The filtrate is magnesium chloride solution
5. To obtain magnesium chloride powder, evaporate the solution till dryness
6. To obtain magnesium chloride crystals, heat the solution while continuously dipping a glass rod in the
solution
7. When you observe crystals starting to form on the glass rod, turn heat off and leave the mixture to cool down
slowly
8. When the crystals are obtained, dry them between two filter papers
Observations of this type of reactions:
 Bubbles of colorless gas evolve (hydrogen). To test approach a lighted splint if hydrogen is present it makes
a pop sound
 The temperature rises (exothermic reaction)
 The metal disappears
You know the reaction is over when:
 No more gas evolves
 No more magnesium can dissolve
 The temperature stops rising
 The solution becomes neutral

4. Proton Donor and Acceptor Theory:
When an acid and a base react, water is formed. The acid gives away an H+ ion and the base accepts it to form
water by bonding it with the OH- ion. A hydrogen ion is also called a proton this is why an acid can be called Proton
Donor and a base can be called Proton Acceptor.
Neutralization Method:
Acis + Base → Salt + Water
Note: This method is used to make salts of metals below hydrogen in the reactivity series. If the base is a metal
oxide or metal hydroxide, the products will be salt and water only. If the base is a metal carbonate, the products will
be salt, water and carbon dioxide.
Type 1:
Acid + Metal Oxide → Salt + Water
To obtain copper sulfate salt given copper oxide and sulfuric acid:
CuO + H2SO4 → CuSO4 + H2O
 Add 100 cm3 of sulfuric acid to a beaker
 Add excess mass of Copper oxide
 When the reaction is over, filter the excess copper oxide off
 The filtrate is a copper sulfate solution, to obtain copper sulfate powder evaporate the solution till dryness
 To obtain copper sulfate crystals, heat the solution white continuously dipping a glass rod in it
 When you observe crystals starting to form on the glass rod, turn heat of and leave the mixture to cool down
slowly
 When you obtain the crystals dry them between two filter papers
Observations of this reaction:
 The amount of copper oxide decreases
 The solution changes color from colorless to blue
 The temperature rises
 You know the reaction is over when
 No more copper oxide dissolves
 The temperature stops rising
 The solution become neutral
Type 2:
Acid + Metal Hydroxide → Salt + Water
to obtain sodium chloride crystals given sodium hydroxide and hydrochloric acid:
HCl + NaOH → NaCl + H2O
 Add 100 cm3of dilute hydrochloric acid to a beaker
 Add excess mass of sodium hydroxide
 When the reaction is over, filter the excess sodium hydroxide off
 The filtrate is sodium chloride solution, to obtain sodium chloride powder, evaporate the solution till dryness
 To obtain sodium chloride crystals, hear the solution while continuously dipping a glass rod in it
 When crystals start to form on the glass rod, turn heat off and leave the mixture to cool down slowly
 When the crystals are obtained, dry them between two filter papers
Observations:
 Sodium hydroxide starts disappearing
 Temperature rises
You know the reaction is over when:
 The temperature stops rising
 No more sodium hydroxide can dissolve
 The pH of the solution becomes neutral
Type 3:

5. Acid + Metal Carbonate → Salt + Water + Carbon Dioxide
To obtain copper sulfate salt given copper carbonate and sulfuric acid:
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
 Add 100 cm3 of dilute sulfuric acid to a beaker
 Add excess mass of copper carbonate
 When the reaction is over, filter excess copper carbonate off
 The filtrate is a copper sulfate solution, to obtain copper sulfate powder evaporate the solution till dryness
 To obtain copper sulfate crystals, heat the solution white continuously dipping a glass rod in it
 When you observe crystals starting to form on the glass rod, turn heat of and leave the mixture to cool down
slowly
 When you obtain the crystals dry them between two filter papers
Observations:
 Bubbles of colorless gas (carbon dioxide) evolve, test by approaching lighted splint, if the CO2 is present
the flame will be put off
 Green Copper carbonate starts to disappear
 The temperature rises
 The solution turns blue
You know the reaction is finished when:
 No more bubbles are evolving
 The temperature stops rising
 No more copper carbonate can dissolve
 The pH of the solution becomes neutral
Titration Method:
This is a method to make a neutralization reaction between a base and an
acid producing a salt without any excess. In this method, the experiment is
preformed twice, the first time is to find the amounts of reactants to use,
and the second experiment is the actual one.
1st Experiment:
 Add 50 cm3 of sodium hydroxide using a pipette to be accurate to
flask
 Add 5 drops of phenolphthalein indicator to the sodium hydroxide.
The solution turns pink indicating presence of a base
 Fill a burette to zero mark with hydrochloric acid
 Add drops of the acid to conical flask
 The pink color of the solution becomes lighter
 When the solution turns colorless, stop adding the acid (End point:
is the point at which every base molecule is neutralized by an acid
molecule)
 Record the amount of hydrochloric acid used and repeat the
experiment without using the indicator
 After the 2nd experiment, you will have a sodium chloride solution.
Evaporate it till dryness to obtain powdered sodium chloride or
crystalize it to obtain sodium chloride crystals
Preparing Insoluble Salts:
Precipitation Method:
A precipitation reaction is a reaction between two soluble salts. The products of a precipitation reaction are two
other salts, one of them is soluble and one is insoluble (precipitate).
Example: To obtain barium sulfate salt given barium chloride and sodium sulfate:
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
Ionic Equation: Ba2+ + SO4
2- → BaSO4
 Add the two salt solutions in a beaker
 When the reaction is over, filter and take the residue

6.  Wash the residue with distilled water and dry it in the oven
Observations:
 Temperature increases
 An insoluble solid precipitate (Barium sulfate) forms
You know the reaction is over when:
 The temperature stops rising
 No more precipitate is being formed
Controlling Soil pH:
If the pH of the soil goes below or above 7, it has to be neutralized using an acid or a base. If the pH of the soil goes
below 7, calcium carbonate (lime stone) is used to neutralize it. The pH of the soil can be measured by taking a
sample from the soil, crushing it, dissolving in water then measuring the pH of the solution.
Colors of Salts:
Salt Formula Solid In Solution
Hydrated copper sulfate CuSO4.5H2O Blue crystals Blue
Anhydrous copper sulfate CuSO4 White powder Blue
Copper nitrate Cu(NO3)2 Blue crystals Blue
Copper chloride CuCl2 Green Green
Copper carbonate CuCO3 Green Insoluble
Copper oxide CuO Black Insoluble
Iron(II) salts E.g.: FeSO4, Fe(NO3)2 Pale green crystals Pale green
Iron(III) salts E.g.: Fe(NO3)3 Reddish brown Reddish brown
Tests for Gases:
Gas Formula Tests
Ammonia NH3 Turns damp red litmus paper blue
Carbon dioxide CO2 Turns limewater milky
Oxygen O2 Relights a glowing splint
Hydrogen H2 ‘Pops’ with a lighted splint
Chlorine Cl2 Bleaches damp litmus paper
Nitrogen dioxide NO2 Turns damp blue litmus paper red
Sulfur dioxide SO2 Turns acidified aqueous potassium dichromate(VI) from orange to green
Tests for Anions:
Anion Test Result
Carbonate (CO32-) Add dilute acid
Effervescence,
carbon dioxide produced
Chloride (Cl-)
(in solution)
Acidify with dilute nitric acid, then add
aqueous silver nitrate
White ppt.
Iodide (I-)
(in solution)
Acidify with dilute nitric acid, then add
aqueous silver nitrate
Yellow ppt.
Nitrate (NO3-)
(in solution)
Add aqueous sodium hydroxide, then
aluminium foil; warm carefully
Ammonia produced
Sulfate (SO42-) Acidify, then add aqueous barium nitrate White ppt.

7. Tests for aqueous cations:
Cation Effect of aqueous sodium hydroxide Effect of aqueous ammonia
Aluminium (Al3+)
White ppt., soluble in excess giving a
colourless solution
White ppt., insoluble in excess
Ammonium (NH4+) Ammonia produced on warming –
Calcium (Ca2+) White ppt., insoluble in excess No ppt. or very slight white ppt.
Copper (Cu2+) Light blue ppt., insoluble in excess
Light blue ppt., soluble in excess,
giving a dark blue solution
Iron(II) (Fe2+) Green ppt., insoluble in excess Green ppt., insoluble in excess
Iron(III) (Fe3+) Red-brown ppt., insoluble in excess Red-brown ppt., insoluble in excess
Zinc (Zn2+)
White ppt., soluble in excess,
giving a colourless solution
White ppt., soluble in excess,
giving a colourless solution

8. Acid-Base Titrations
Acid-Base titrations are usually used to find the the amount of a known acidic or basic substance through acid base reactions. The
analyte (titrand) is the solution with an unknown molarity. The reagent (titrant) is the solution with a known molarity that will react with
the analyte.
Procedure
The analyte is prepared by dissolving the substance being studied into a solution.The solution is usually placed in a flask for titration. A
small amount of indicator is then added into the flask along with the analyte. The reagent is usually placed in a burette and slowly
added to the analyte and indicator mixture.The amountof reagentused is recorded when the indicator causes a change in the color of
the solution.
Some titrations requires the solution to be boiled due to the CO2 created from the acid-base reaction. The CO2 forms carboinic
acid (H2CO3) when dissolved in water. The carbonic acid then acts as a buffer, reducing the accuracy of data. After boiling most of the
CO2will be removed from the solution allowing the solution to be titrated to a more accurate endpoint. The endpoint is the point where
all of the analyte has be reacted with the reagent.
Choosing an Indicator
A useful indicator has a strong color that changes quickly near its pKa. These traits are desirable so only a small amount of an
indicator is needed.If a large amountofindicator is used, the indicator will effect the final pH, lowering the accuracy of the experiment.
The indicator should also have a pKa value near the pH of the titration's endpoint. For example a analyte that is a weak base would
require an indicator with a pKa less than 7.Choosing an indicator with a pKa near the endpoint's pH will also reduce error b ecause the
color change occurs sharply during the endpoint where the pH spikes, giving a more precise endpoint.
Figure 1: A Basic Titration Curve
Notice that this reaction is between a weak acid and a strong base so phenolphthalein with a pKa of 9.1 would be a better cho ice than
methyl orange with a pKa of 3.8. If in this reaction we were to use methyl orange as the indicator color changes would occur all
throughout the region highlighted in pink. The data obtained would be hard to determine due to the large range of color change, and
inaccurate as the color change does not even lie with the endpoint region. Phenolphthalein on the other hand changes color ra pidly
near the endpoint allowing for more accurate data to be gathered.
Calculations
Multiply the volume of reagentadded to get to the endpoint, with the molarity of the reagent to find the moles of reagent used. With the
balanced equation of the acid-base reaction in question to find the moles of unknown substance. Then the original molarity can be
calculated by dividing through with the initial volume.
For example an unknown molarity of HCl acts as the analyte. 50mL of it is placed into a flask and a 0.1M solution of NaOH wil l be the
reagent.The endpoint's pH is 7 so litmus,with a pKa of 6.5 is chosen. The color of the solution changes when 10mL of 0.1M NaOH is
added.
The balanced chemical equation related to is
HCl(aq)+NaOH(aq)→H2O(l)+Na++Cl−
Or justthe net ionic equation
H++OH−→H2O(l)
The following equation can then be derived
X = 0.0010 mol of HCl
The molarityis now easilysolved for
0.0010 mol HCl / 0.050 L = 0.020M HCl

9. Buffers
A buffer is a solution thatcan resistpH change upon the addition ofan acidic or basic components.It is able to neutralize small
amounts ofadded acid or base, thus maintaining the pH of the solution relatively stable.This is importantfor processes and/or
reactions which require specific and stable pH ranges.Buffer solutions have a working pH range and capacity which dictate how much
acid/base can be neutralized before pH changes,and the amountby which it will change.
What is a buffer composed of?
To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its
conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when
preparing the buffer. For example, the following could function as buffers when together in solution:
 Acetic acid (weak organic acid w/ formula CH3COOH) and a salt containing its conjugate base, the acetate anion (CH3COO-
), such as
sodium acetate (CH3COONa)
 Pyridine (weak base w/ formula C5H5N) and a salt containing its conjugate acid, the pyridinium cation (C5H5NH+
), such as Pyridinium
Chloride.
 Ammonia (weak base w/formula NH3) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide
(NH4OH)
How does a buffer work?
A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable
amounts atequilibrium and are able to neutralize small amounts of other acids and bases (in the form of H 3O+
and OH-
) when the are
added to the solution.
In order to clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer.
Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F-
ion and solvated protons (H3O+
), which does
not allow it to dissociate completelyin water. Therefore,if we obtain HF in an aqueous solution, we establish the following equilibrium
with only slight dissociation (Ka(HF) = 6.6x10-4
, strongly favors reactants):
HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)
We can then add and dissolve sodium fluoride into the solution and mixthe two until we reach the desired volume and pH at which we
want to buffer.
When Sodium Fluoride dissolves in water,the reaction goes to completion,thus we obtain:
NaF(aq)+H2O(l)→Na+(aq)+F−(aq)
Since Na+
is the conjugate ofa strong base,it will have no effect on the pH or reactivity of the buffer. The addition ofNaF to the s olution
will, however, increase the concentration of F-
in the buffer solution, and, consequently, by Le Chatlier's principle, lead to slightly less
dissociation ofthe HF in the previous equilibrium,as well. The presence ofsignificantamounts of both the conjugate acid, HF, and the
conjugate base, F-
, allows the solution to function as a buffer.
This buffering action can be seen in the titration curve of a buffer solution.
As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering
capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted
throughneutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.
If acid is added:
F−(aq)+H3O+(aq)⇌HF(aq)+H2O(l)
In this reaction, the conjugate base,F-
, will neutralize the added acid, H3O+
, and this reaction goes to completion, because the reaction
of F-
with H3O+
has an equilibrium constant much greater than one. (In fact, the equilibrium constant the reaction as written is just the

10. inverse of the Ka for HF: 1/Ka(HF) = 1/(6.6x10-4
) = 1.5x10+3
.) So long as there is more F-
than H3O+
, almost all of the H3O+
will be
consumed and the equilibrium will shiftto the right, slightlyincreasing the concentration ofHF and slightlydecreasing the concentration
of F-
, but resulting in hardly any change in the amount of H3O+
present once equilibrium is re-established.
If base is added:
HF(aq)+OH−(aq)⇌F−(aq)+H2O(l)
In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH -
, and the equilibrium will again shift to the right,
slightly increasing the concentration of F-
in the solution and decreasing the amount of HF slightly. Again, since most of the OH -
is
neutralized, little pH change will occur.
These two reactions can continue to alternate back and forth with little pH change.
Selecting proper components for desired pH
Buffers function bestwhen the pKa of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to
be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10).
For example, we know the Ka for hydroflouric acid is 6.6 x 10-4
so its pKa= -log(6.6 x 10-4
) = 3.18. So, a hydrofluoric acid buffer would
work best in a buffer range of around pH = 3.18.
For the weak base ammonia (NH3), the value of Kb is 1.8x10-5
, implying that the Ka for the dissociation of its conjugate acid, NH4
+
, is
Kw/Kb=10-14
/1.8x10-5
= 5.6x10-10
. Thus, the pKa for NH4
+
= 9.25, so buffers using NH4
+
/NH3 will work best around a pH of 9.25. (It's always
the pKa of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pK b of the
conjugate base, obviously.)
When the desired pH ofa buffer solution is near the pKa of the conjugate acid being used (i.e., when the amounts ofconjugate acid and
conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a
simple approximation of the solution pH, as we will see in the next section.
Example: HF Buffer
In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of
3.0. We can use the Henderson-Hasselbalch equation to calculate the necessary ratio of F-
and HF.
pH=pKa+log[Base][Acid]
3.0=3.18+log[Base][Acid]
log[Base][Acid]=−0.18
[Base][Acid]=10−0.18
[Base][Acid]=0.66
This is simplythe ratio of the concentrations ofconjugate base and conjugate acid we will need in our solution.However, wh at if we have
100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in ord er to create a
buffer at said pH (3.0)?
We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66. From a table of molar
masses,such as a periodic table,we can calculate the molar mass ofNaF to be equal to 41.99 g/mol.HF is a weak acid with a Ka = 6.6 x
10-4
and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F-
we need to add.
The dissociation reaction is:
HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)
We could use ICE tables to calculate the concentration of F-
from HF dissociation, but, since Ka is so small, we can approximate that
virtually all of the HF will remain undissociated,so the amountofF-
in the solution from HF dissociation will be negligible. Thus, the [HF]
is about 1 M and the [F-
] is close to 0. This will be especially true once we have added more F-
, the addition of which will even further
suppress the dissociation of HF.
We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66. Thus, [F-
] should be about 0.66 M. For 100 mL of solution,
then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F-
. Since we are adding NaF as our source of F-
, and since NaF completely
dissociates in water, we need 0.066 moles of NaF. Thus, 0.066 moles x 41.99 g/mol = 2.767 g.
Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of
"Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "mole s of base
over moles ofacid." The pH of the solution does not,it turns out, depend on the volume! (This is only true so long as the solution does
not get so dilute that the autoionization of water becomes an important source of H +
or OH-
. Such dilute solutions are rarely used as
buffers, however.)
Adding Strong Acids or Bases to Buffer Solutions
Now that we have this nice F-
/HF buffer, let's see what happens when we add strong acid or base to it. Recall that the amount of F-
in
the solution is 0.66M x 0.1 L = 0.066 moles and the amount of HF is 1.0 M x 0.1L = 0.10 moles. Let's double check the pH usin g
theHenderson-Hasselbalch Approximation, but using moles instead of concentrations:

11. pH = pKa + log(Base/Acid) = 3.18 + log(0.066 moles F-
/0.10 moles HF) = 3.00
Good. Now let's see what happens when we add a small amount of strong acid, such as HCl. When we put HCl into water, it
completely dissociates into H3O+
and Cl-
. The Cl-
is the conjugate base of a strong acid so is inert and doesn't affect pH, and we can
just ignore it. However, the H3O+
can affect pH and it can also react with our buffer components. In fact, we already discussed what
happens. The equation is:
F−(aq)+H3O+(aq)⇌HF(aq)+H2O(l)
For every mole of H3O+
added,an equivalentamountofthe conjugate base (in this case,F-
) will also react,and the equilibrium constant
for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F-
is used up before reacting
away all of the H3O+
, then the remaining H3O+
will affect the pH directly. In this case,the capacity of the buffer will have been exceeded -
a situation one tries to avoid. However, for our example, let's say that the amount of added H 3O+
is smaller than the amount of F-
present, so our buffer capacity is NOT exceeded. For the purposes of this example, we'll let the added H3O+
be equal to 0.01 moles
(from 0.01 moles ofHCl).Now, if we add 0.01 moles ofHCl to 100 mL of pure water, we would expect the pH of the resulting s olution to
be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0).
However, we are adding the H3O+
to a solution that has F-
in it, so the H3O+
will all be consumed by reaction with F-
. In the process, the
0.066 moles of F-
is reduced:
0.066 initial moles F-
- 0.010 moles reacted with H3O+
= 0.056 moles F-
remaining
Also during this process,more HFis formed by the reaction:
0.10 initial moles HF + 0.010 moles from reaction of F-
with H3O+
= 0.11 moles HF after reaction
Plugging these new values into Henderson-Hasselbalch gives:
pH = pKa + log (base/acid) = 3.18 + log (0.056 moles F-
/0.11 moles HF) = 2.89
Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of
strong acid.

12. Acid and Base Indicators
The most common method to get an idea about the pH of solution is to use an acid base indicator. An indicator is a large orga nic
molecule that works somewhat like a " color dye". Whereas most dyes do not change color with the amount of acid or base prese nt,
there are many molecules,known as acid- base indicators , which do respond to a change in the hydrogen ion concentration. Most of
the indicators are themselves weak acids.
Indicators
The mostcommon indicator is found on "litmus"paper.It is red below pH 4.5 and blue above pH 8.2.
Color Blue Litmus Red Litmus
Acid turns red stays same
Base stays same turns blue
Other commercial pH papers are able to give colors for every main pH unit. Universal Indicator, which is a solution of a mixt ure of
indicators is able to also provide a full range of colors for the pH scale.
A variety of indicators change color at various pH levels. A properly selected acid-base indicator can be used to visually "indicate" the
approximate pH of a sample.An indicator is usuallysome weak organic acid or base dye that changes colors atdefinite pH values. The
weak acid form (HIn) will have one color and the weak acid negative ion (In-
) will have a different color. The weak acid equilibrium is:
HIn → H+
+ In-
 For phenolphthalein: pH 8.2 = colorless; pH 10 = red
 For bromophenol blue:pH 3 = yellow; pH 4.6 = blue
See the graphic below for colors and pH ranges.
Magic Pitcher Demonstration
Phenolphthalein is an indicator of acids (colorless) and bases (pink). Sodium hydroxide is a base, and it was in the pitcher at the
beginning, so when added to the phenolphthalein in beakers 2 and 4, it turned pink (top half of the graphic).
Reaction
Equilibrium: HIn → H+
+ In-
colorless red
The equilibrium shifts right,HIn decreases,and In-
increases.As the pH increase between 8.2 to 10.0 the color becomes red because
of the equilibrium shifts to form mostly In-
ions.

13.  The third beaker has only the NaOH but no phenolphthalein, so it remained colorless. The first beaker contain acetic acid and is
skipped over at first.
 After pouring beakers 2, 3, 4 back into the pitcher it give a pink solution.
 Bottom half of the graphic: When the pitcher is then poured back into beakers 2, 3, 4 it is a pink solution.
 In the first beaker,a strange thing happens in that the pink solution coming outofthe pitcher now changes to colorless.Th is happens
because the first beaker contains some vinegar or acetic acid which neutralizes the NaOH, and changes the solution from basic to
acidic. Under acidic conditions, the phenolphthalein indicator is colorless.
Neutralization reaction:
HC2H3O2 + NaOH → Na(C2H3O2) + HOH
Explain the color indicator change
Use equilibrium principles to explain the color change for phenolphthalein atthe end of the demonstration.
Solution
The simplified reaction is:H+
+ OH-
→ HOH
As OH-
ions are added, they are consumed by the excess of acid already in the beaker as expressed in the above equation. The
hydroxide ions keep decreasing and the hydrogen ions increase, pH decreases.
See lower equation:The indicator equilibrium shifts left,In-
ions decrease.Below pH 8.2 the indicator is colorless.As H+
ions are further
increased and pH decreases to pH 4-5, the indicator equilibrium is effected and changes to the colorless HIn form.
Equilibrium: HIn → H+
+ In-
colorless red

14. Molecular Basis for the Indicator Color Change
Color changes in molecules can be caused by changes in electron confinement. More confinement makes the light absorbed more
blue, and less makes it more red.
How are electrons confined in phenolphthalein? There are three benzene rings in the molecule. Every ato m involved in a double bond
has a p orbital which can overlap side-to-side with similar atoms nextto it. The overlap creates a 'pi bond' which allows the electrons in
the p orbital to be found on either bonded atom.These electrons can spread like a cloud over any region of the molecule that is flat and
has alternating double and single bonds. Each of the benzene rings is such a system.
See the far left graphic - The carbon atom at the center (adjacent to the yellow circled red oxygen atom) doesn't have a p-orbital
available for pi-bonding, and it confines the pi electrons to the rings. The molecule absorbs in the ultraviolet, and this form of
phenolphthalein is colorless.
In basic solution, the molecule loses one hydrogen ion. Almost instantly, the five-sided ring in the center opens and the electronic
structure around the center carbon changes (yellow circled atoms) to a double bond which now does contain pi electrons. The p i
electrons are no longer confined separately to the three benzene rings, but because of the change in geometry around the yellow
circled atoms, the whole molecule is now flat and electrons are free to move within the entire molecule. The result of all of these
changes is the change in color to pink. Chime: Phenolphthalein
Many other indicators behave on the molecular level in a similar fashion (the details may be different) but the result is a c hange in
electronic structure along with the removal of a hydrogen ion from the molecule. Plant pigments in flowers and leaves also behave in
this fashion.

15. Brønsted Concept of Acids and Bases
Table of Contents
1. 1. Brønsted-LoweryDefinition
2. 2. Acids are Proton Donors and Bases are Proton Acceptors
3. 3. Questions
4. 4. Outside Links
5. 5. Sources
6. 6. Contributors
In 1923, chemists Johannes Brønsted and Martin Lowry independentlydeveloped definitions ofacids and bases based on compound s
abilities to either donate or accept protons (H+
ions). Here, acids are defined as being able to donate protons in the form of hydrogen
ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no
longer required to be present in the solution for acid and base reactions to occur.
Brønsted-Lowery Definition
In 1923, J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid -base
reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by
whether the molecule contains hydrogen and hydroxide ion is too limiting.The main effect of the Brønsted -Lowry definition is to identify
the proton (H+
) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:
HA + Z ↔ A-
+ HZ+
Acid Base
Donates hydrogen ions Accepts hydrogen ions.
HCl+
HOH → H3O+
+ Cl-
HOH+
NH3→ NH4
+
+ OH-
The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the
HOH it is a base in the first case and an acid in the second case.
To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the
number ofhydrogens has decreased thatsubstance is the acid (donates hydrogen ions).If the number ofhydrogens has increased that
substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is
viewed in reverse a new acid and base can be identified.The substances on the right side of the equation are called conjugate acid and
conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.
Acids are Proton Donors and Bases are Proton
Acceptors
So what does this mean? For a reaction to be in equilibrium a transfer ofelectrons needs to occur. The acid will give an electron away
and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up
ofconjugate acids and conjugate bases.
HA+Z⇌A−+HZ+
A stands for an Acidic compound and Z stands for a Basic compound

16.  A Donates H to form HZ+
.
 Z Accepts H from A which forms HZ+
 A-
becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibr ium
 HZ+
becomes a conjugate acid of Z and in the reverse reaction it donates a H to A-
recreating Z in order to remain in equilibrium
Brønsted Concept of Acids and Bases
Table of Contents
1. 1. Brønsted-LoweryDefinition
2. 2. Acids are Proton Donors and Bases are Proton Acceptors
3. 3. Questions
4. 4. Outside Links
5. 5. Sources
6. 6. Contributors
In 1923, chemists Johannes Brønsted and Martin Lowry independentlydeveloped definitions ofacids and bases based on compound s
abilities to either donate or accept protons (H+
ions). Here, acids are defined as being able to donate protons in the form of hydrogen
ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no
longer required to be present in the solution for acid and base reactions to occur.
Brønsted-Lowery Definition
In 1923, J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base
reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by
whether the molecule contains hydrogen and hydroxide ion is too limiting.The main effect of the Brønsted-Lowry definition is to identify
the proton (H+
) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:
HA + Z ↔ A-
+ HZ+
Acid Base
Donates hydrogen ions Accepts hydrogen ions.
HCl+
HOH → H3O+
+ Cl-
HOH+
NH3→ NH4
+
+ OH-
The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the
HOH it is a base in the first case and an acid in the second case.
To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the
number ofhydrogens has decreased thatsubstance is the acid (donates hydrogen ions).If the number ofhydrogens has increased that
substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is
viewed in reverse a new acid and base can be identified.The substances on the right side of the equation are called conjugate acid and
conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.
Acids are Proton Donors and Bases are Proton
Acceptors

17. So what does this mean? For a reaction to be in equilibrium a transfer ofelectrons needs to occur. The acid will give an electron away
and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up
ofconjugate acids and conjugate bases.
HA+Z⇌A−+HZ+
A stands for an Acidic compound and Z stands for a Basic compound
 A Donates H to form HZ+
.
 Z Accepts H from A which forms HZ+
 A-
becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibr ium
 HZ+
becomes a conjugate acid of Z and in the reverse reaction it donates a H to A-
recreating Z in order to remain in equilibrium

18. Lewis Concept of Acids and Bases
Acids and bases are an important part of chemistry. One of the most applicable theories is the Lewis acid/base motif that extends the
definition of an acid and base beyond H+
and OH-
ions as described by Brønsted-Lowry acids and bases.
Introduction
The Brønsted acid-base theory has been used throughout the history of acid and base chemistry. However, this theory is very
restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the
theory doesn't necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to
describe acids and bases. His theory gave a generalized explanation ofacids and bases based on structure and bonding. Through the
use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory
used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base
donates an electron pair.
Figure 1 Above: A Lewis Base (B) donates itelectrons to a Lewis Acid (A) resulting in a coordinate covalently bonded compoun d, also
known as an adduct.
The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond, as shown in Figure 1 above. A coordina te
covalent bond is justa type of covalent bond in which one reactantgives it electron pair to another reactant. In this case the lewis base
donates its electrons to the lewis acid. When they do react this way the resulting product is called an addition compound, or more
commonly an adduct.
 Lewis Acid: a species thataccepts an electron pair (i.e., an electrophile) and will have vacant orbitals
 Lewis Base:a species thatdonates an electron pair (i.e., a nucleophile) and will have lone-pair electrons
Lewis Acids
Lewis acids acceptan electron pair. Lewis Acids are Electrophilic meaning thatthey are electron attracting. When bonding with a base
the acid uses its lowestunoccupied molecular orbital or LUMO (see Fig. 2).
 Various species can act as Lewis acids. All cations are Lewis acids since they are able to accept electrons. (e.g., Cu2+
, Fe2+
, Fe3+
)
 An atom, ion, or molecule with an incomplete octet of electrons can act as an Lewis acid (e.g., BF3, AlF3).
 Molecules where the central atom can have more than 8 valence shell electrons can be electron acceptors, and thus are classified as
Lewis acids (e.g., SiBr4, SiF4).
 Molecules that have multiple bonds between two atoms of different electronegativities (e.g., CO2, SO2)
Lewis Bases
Lewis Bases donate an electron pair. Lewis Bases are Nucleophilic meaning that they “attack” a positive charge with their lone pair.
They utilize the highest occupied molecular orbital or HOMO (see Fig. 2). An atom, ion, or molecule with a lone-pair of electrons can
thus be a Lewis base. Each of the following anions can "give up" their electrons to an acid.
Example:OH-
, CN-
, CH3COO-
, :NH3, H2O:, CO:
HOMO & LUMO
Lewis base's HOMO(highestoccupied molecular orbital) interacts with the Lewis acid's LUMO(lowestunoccupied molecular orbi tal) to
create bonded molecular orbitals.Both Lewis Acids and Bases contain HOMO and LUMOs but only the HOMO is considered for Bases
and only the LUMO is considered for Acids. (see fig. 2 below)
Figure 2 Above: Lewis Acids have vacant orbitals so they are in a lower energy level. While lewis bases have lone pair electrons to
share and thus occupy a higher energy level.

19. Complex Ion / Coordination Compounds
Complex ions are polyatomic ions, which are formed from a central metal ion that has other smaller ions joined around it. While
Brønsted theory can't explain this reaction Lewis acid-base theory can help. A Lewis Base is often the ligand of a coordination
compound with the metal acting as the Lewis Acid (see Oxidation States of Transition Metals).
Al3++6H2O⇌[Al(H2O)6]3+
The aluminum ion is the metal and is a cation with an unfilled valence shell,and it is a Lewis Acid. Water has lone-pair electrons and is
an anion,thus it is a Lewis Base.
Figure 3 (Left): Aluminum ion acts as a lewis acid and accepts the electrons from water, which is acting as a lewis base. This helps
explain the resulting hexaaquaaluminum(III) ion.
The Lewis Acid accepts the electrons from the Lewis Base which donates the electrons. Another case where Lewis acid -base theory
can explain the resulting compound is the reaction of ammonia with Zn2+
.
Zn2++4NH3→[Zn(NH3)4]4+
Similarly, the Lewis Acid is the zinc Ion and the Lewis Base is NH3. Note how Brønsted Theory of Acids and Bases will not be able to
explain how this reaction occurs because there are no H+ or OH− ions involved. Thus, Lewis Acid and Base Theory allows us to
explain the formation of other species and complex ions which do not ordinarily contain hydronium or hydroxide ions. One is a ble to
expand the definition of an acid and a base via the Lewis Acid and Base Theory. The lack of H+ or OH− ions in many complex ions
can make it harder to identify which species is an acid and which is a base. Therefore, by defining a species that donates an electron
pair and a species that accepts an electron pair, the definition of a acid and base is expanded.
Amphoterism
As of now you should know thatacids and bases are distinguished as two separate things however some substances can be both a n
acid and a base.You may have noticed this with water, which can act as both an acid or a base.This ability of water to do this makes it
an amphoteric molecule. Water can act as an acid by donating its proton to the base and thus becoming its conjugate acid, OH -.
However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H3O+
.
 Water acting as an Acid:
H2O+NH3→NH+4+OH−
 Water acting as a Base:
H2O+HCl→Cl−+H3O+
You may have noticed that the degree to which a molecule acts depends on the medium in which the molecule has been placed in.
Water does not act as an acid in an acid medium and does notactas a base in a basic medium.Thus,the medium which a molecu le is
placed in has an effect on the properties of that molecule. Other molecules can also act as either an acid or a base. For exa mple,
Al(OH)3+3H+→Al3++3H2O
 where Al(OH)3 is acting as a Lewis Base.
Al(OH)3+OH−→Al(OH)−4
 where Al(OH)3 is acting as an Lewis Acid.
Note how the amphoteric properties ofthe Al(OH)3 depends on whattype of environmentthat molecule has been placed in
Arrhenius Concept of Acids
and Bases
The Arrhenius definition can be summarized as "Arrhenius acids form hydrogen ions in
aqueous solution with Arrhenius bases forming hydroxide ions."

20. Introduction
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of
compounds, termed acids and bases. When dissolved in an aqueous solution, certain
ions were released into the solution. As defined by Arrhenius, acid-base reactions are
characterized by acids, which dissociate in aqueous solution to form hydrogen ions (H+) and
bases, which form hydroxide (OH−) ions.
Acids are defined as a compound or element that releases hydrogen (H+) ions into the
solution.
In this reaction nitric acid (HNO3) disassociates into hydrogen (H+) and nitrate (NO3-) ions
when dissolved in water.
Bases are defined as a compound or element that releases hydroxide (OH-) ions into the
solution.
In this reaction lithium hydroxide (LiOH) dissociates into lithium (Li+) and hydroxide (OH-) ions
when dissolved in water.

21. Definition of pH, pOH, "p", sample calculations
General
The acid potential of aqueous solutions is measured in terms of the pH scale. The symbol "p" means
take the negative logarithm of whatever follows in the formula. for pH, pOH, p[anything] . The pH scale
is intended to be a convenience. Tremendous swings in hydrogen ion (hydronium ion) concentration
occur in water when acids or bases are mixed with water. These changes can be as big as 1 x 1014,
This means concentrations can change by multiples as big as one hundred trillion,
100,000,000,000,000.
The pH scale is a logarithmic scale. Every multiple of ten in H1+ concentration equals one unit on the
logarithm scale. Physically the pH is intended to tell what the acid "potential" is for a solution.
In a sense the system is INVERTED so a low pH value indicates a great acid potential while a high pH
indicates a low acid potential. ( Sad but true this is upside down and counter intuitive.) The pH values
range from negative values to number above 14. Commonly the scale is often misrepresented as
ranging from 0 to 14. We will see that negative values are possible.
Definition of pH, pOH, pKw, pKa, pKb
The p" factor" is defined as the log of the whatever quantity that follows the symbol. The "p" is an
operator. It communicates the instruction to calculate the negative log of any quantity that follows the
symbol. The definition of pH in equation form is
pH = -log[H1+] where [H1+] means the molar concentration of hydronium ions, M = moles / liter
This allows the definition of the following series of quantities.
pOH = -
log[OH-]
the negative log of the hydroxide ion molarity
pKw = -log Kw the negative log of the water ion product , Kw
pKa = -log Ka the negative log of the acid dissociation constant, Ka
pKb = -log Kb the negative log of the base dissociation constant, Kb
The relationship pH + pOH = 14
In a water solution the ion product for water is:
[H+] [OH-] = Kw = 1 X 10-14
Take the -log of both sides of the equation
- log [H+] +(- log [OH- ]) = - log [1 X 10-14 ]
pH + pOH = 14
Calculations of pH
For strong acids like HCl the molar concentrations are essentially the hydronium ion concentration.
These strong acids can produce solutions where the pH can be equal to or less than 1, the pH value

22. would have a value from 0-14.
Example: Determination of pH from [H3O+]
What is the pH of a solution whose [H3O+] = 1 X 10-4 M
pH = -log[H3O+]
pH = - log[1 X 10-4]
pH = - [ log 1 + log 10-4 ]
Note: When you multiply numbers you always ADD their log forms
log 1 is always zero
log 10x = x so log 10-4 = -4
pH = - [ log 1 + log 10-4 ] = - [ 0 + (-4) ] = - [-4 ] = +4
Example: Determination of pH from [H3O+] when coefficient is other than "1"
What is the pH of a solution whose [H3O+] = 2.5 X 10 -5 M
pH = -log[H3O+]
pH = - log[2.5 X 10 -5]
pH = - [ log 2.5 + log 10-5 ]
Note: When you multiply numbers you always ADD their log forms
log 10x = x so log 10-5 = -5
log 2.5 can be determined using a calculator having the log
function key:
Enter the number in this case 2.5
depress the log key
Read the display which should be .3979 for this problem
pH = - [.3979 - 5] = 4.6021 or +4.602
Alternately if you can enter a number in scientific notation into your calculator key in 2.5 X
10 -5
depress the log key
Read the display which should be -4.602 for this problem
Multiply by -1 to get + 4.602
Example: Determination of pH from [OH1- ] using defintion pOH and equation pH + pOH = 14
Calculate the pH of a solution that has a [OH1-] = 1 X 10-5 M
Determine pOH = -log[OH1- ] = -log [1 X 10-5 ] = 5
Use the relationship pH + pOH = 14
pH + 5 = 14
pH = 14 -5 = 9

23. Hydrolysis of salts
Return to the Acid Base menu
A Brief Introduction to Hydrolysis Calculations
Hydrolysis happens when a substance chemically reacts with water. Hydrolysis should be
distinguished from solvation, which is the process of water molecules associating
themselves with individual solute molecules or ions.
I. Salts of Weak Acids
In general, all salts of weak acids behave the same, therefore we can use a generic salt to
represent all salts of weak acids. Let NaA be a generic salt of a weak acid and A¯ its
anion. Here are two specific examples of salts of weak acids:
Substance Formula The anion portion (A¯)
sodium acetate NaC2H3O2 C2H3O2¯
sodium benzoate C6H5COONa C6H5COO¯
By the way, the potassium ion, K+
, (and several others) could also be used above without
affecting any discussions of this topic. As a practical matter, only Na+
and K+
tend to get
used in examples.
The generic chemical reaction (in net ionic form) for hydrolysis may be written thusly:
A¯ + H2O --> HA + OH¯
This reaction is of a salt of a weak acid (NOT the acid) undergoing hydrolysis, the name
for a chemical reaction with water. The salt is NaAc and it is reacting with the water. Keep
in mind that the acid (HAc) does not undergo hydrolysis, the salt does.
It is very important that you notice several things:
1) The Na+
(notice only OH¯ is written) IS NOT involved. Its source is the salt (NaA) that is
dissolving in the water and it DOES NOT affect the pH. Its presence in both writing the
chemical reactions and doing the calculations is deleted. However, keep in mind that
Na+
is present in the solution. Some teacher might want to ask a "sneaky" question on a
test.
2) HA is the UNDISSOCIATED acid. Keep in mind that it is not the acid that makes the
acidic pH of a solution, it is the amount of hydrogen ion (or hydronium ion, H3O+
, if you
wish). In order to produce the hydrogen ion, the acid must dissociate.
3) There is free hydroxide ion (OH¯) in the solution!! This is the thing that makes the pH
greater than 7.
Now, I can see a question forming in your mind. If there is acid (HA) and base (OH¯), why
don't they just react and give back the reactants on the left side? Now, that really is a good
question.
The answer? This reaction is an equilibrium. Now, if you are taking chemistry for the first
time, you probably just got done with equilibrium a few weeks ago and it might have been
hard to understand. That's understandable, but please realize that equilibrium is one of
more important concepts in chemistry. Keep up the work!!
When a chemical reaction comes to equilibrium, there is a mixture of all involved
substances in the reaction vessel. This mixture is characterized by a constant
composition. (Keep in mind that constant composition DOES NOT imply equal

24. composition.) The key point that makes a reaction come to equilibrium is that it is
reversible. This means that both the forward reaction and the reverse reaction can
happen, althought NOT initially with equal probability. The reaction comes to equilibrium
when the rates of the two reactions (forward and reverse) become equal.
So, while it is true that the HA and OH¯ will react in the reverse direction, so can the A¯
and the H2O in the forward direction. The key point is that the reaction happens in such a
way that a small amount (as opposed to zero) of HA and OH¯ are present at equilibrium.
When calculations are done, the important points will be (1) how much OH¯ is formed and
(2) what is the pH of the solution?
Quick answers: (1) the amount of OH¯ formed will be greater than the 10¯7
M value
present in pure water and (2) the pH will be greater than 7, so the solution of the salt of a
weak acid will be basic.
II. Salts of Weak Bases
In general, all salts of weak bases behave the same, therefore we can use a generic salt
to represent all salts of weak bases. Let B be a generic base and HB+
its salt. (Compare
how this is worded compared to the "salt of weak acid" discussion.) HB+
is a cation, but
that word is not used as much in discussions as is "anion" is above. Here are two specific
examples of salts of weak bases:
Substance Formula The cation portion (HB+)
ammonium chloride NH4Cl NH4+
methyl ammonium nitrate CH3NH3NO3 CH3NH3+
The notation HB+
might be a bit confusing. Think of NH4
+
this way:
HNH3
+
NH3 is the base (symbolized by B) and an H+
has been attached to it in a chemical
reaction. the NH3 has been protonated and the result (NH4
+
) is now an acid. Why?
Because it now has a proton to donate.
By the way, the chloride ion, Cl¯, and the nitrate ion, NO3¯ tend to be used in examples.
Other anions of strong acids could be used, but their use is fairly uncommon.
The generic chemical reaction (in net ionic form) for hydrolysis reaction may be written
thusly:
HB+
+ H2O --> B + H3O+
This reaction is of a salt of a weak base (NOT the base) undergoing hydrolysis, the
reaction with water. The salt in this case is HB+
Cl¯ and it is reacting with the water.
Remember, the most common specific example would be ammonium chloride (NH4
+
Cl¯).
Keep in mind that the base (generic example = B, specific example = ammonia or NH3)
does not undergo hydrolysis, the salt does.
It is very important that you notice several things:
1) There is an anion involved, but it is usually not written. For example Cl¯ could be the
anion, but it IS NOT involved. Its source is the salt (HB+
Cl¯) that is dissolving in the water
and it DOES NOT affect the pH. Its presence in writing the appropriate chemical reactions
and doing the calculations is deleted. However, keep in mind that Cl¯ is present in the
solution. Some teacher might want to ask a "sneaky" question on a test.

25. 2) B is the UNPROTONATED base. Keep in mind that it is not the base that makes the
basic pH of a solution, it is the amount of hydroxide ion (OH¯). In order to produce it, the
base must protonated by the water.
3) There is free hydronium ion (H3O+
) in the solution!! This is the thing that makes the pH
less than 7.
Now, I can see a question forming in your mind. If there is base (B) and acid (H3O+
), why
don't they just react and give back the reactants on the left side? Now, that really is a good
question.
The answer, of course, is given in above in the discussion of salts of weak acids. It would
be the same explanation here, so I won't repeat it. What you might want to do, however, is
look at the different phrasing in part I as compared to part II.
Of course, when calculations are done, the important points will be (1) how much H3O+
is
formed and (2) what is the pH of the solution?
Quick answers: (1) the amount of H3O+
formed will be greater than the 10¯7
M value
present in pure water and (2) the pH will be less than 7, so the solution of the salt of a
weak base will be acidic.

26. salt
Characteristics and Classification of Salts
The most familiar salt is sodium chloride, the principal component of common table salt. Sodium chloride, NaCl, and
water, H2O, are formed by neutralization of sodium hydroxide, NaOH, a base, with hydrogen chloride, HCl, an acid:
HCl+NaOH→NaCl+H2O. Most salts are ionic compounds (see chemical bond); they are made up of ions rather than
molecules. The chemical formula for an ionic salt is an empirical formula; it does not represent a molecule but shows the
proportion of atoms of the elements that make up the salt. The formula for sodium chloride, NaCl, indicates that equal
numbers of sodium and chlorine atoms combine to form the salt. In the reaction of sodium with chlorine, each sodium
atom loses an electron, becoming positively charged, and each chlorine atom gains an electron, becoming negatively
charged (see oxidation and reduction); there are equal numbers of positively charged sodium ions and negatively charged
chloride ions in sodium chloride. The ions in a solid salt are usually arranged in a definite crystalline structure, each
positive ion being associated with a fixed number of negative ions, and vice versa.
A salt that has neither hydrogen (H) nor hydroxyl (OH) in its formula, e.g., sodium chloride (NaCl), is called a normal salt.
A salt that has hydrogen in its formula, e.g., sodium bicarbonate (NaHCO3), is called an acid salt. A salt that has hydroxyl
in its formula, e.g., basic lead nitrate (Pb[OH]NO3), is called a basic salt. Since a salt may react with a solvent to yield
different ions than were present in the salt (see hydrolysis), a solution of a normal salt may be acidic or basic; e.g.,
trisodium phosphate, Na3PO4, dissolves in and reacts with water to form a basic solution.
In addition to being classified as normal, acid, or basic, salts are categorized as simple salts, double salts, or complex
salts. Simple salts, e.g., sodium chloride, contain only one kind of positive ion (other than the hydrogen ion in acid salts).
Double salts contain two different positive ions, e.g., the mineral dolomite, or calcium magnesium carbonate,
CaMg(CO3)2. Alums are a special kind of double salt. Complex salts, e.g., potassium ferricyanide, K3Fe(CN)6, contain a
complex ion that does not dissociate in solution. A hydrate is a salt that includes water in its solid crystalline form;
Glauber's salt and Epsom salts are hydrates.
Salts are often grouped according to the negative ion they contain,
e.g., bicarbonate or carbonate, chlorate, chloride, cyanide, fulminate, nitrate, phosphate, silicate, sulfate, or sulfide.