Chemistry Chapter 16 acid-base titration and ph

Acid-base titration and ph
Chapter 16

Aqueous Solutions and the Concept of pH
Section 1

Hydronium and Hydroxide Ions
Acids and bases form hydroxide and hydronium ions
These ions are not the only ones in an aqueous solution
Hydronium and hydroxide and also produced by water

Self-Ionization of Water
Water can undergo self-ionization
Self-ionization of water two water molecules make a hydronium and hydroxide ion by transfer of a proton

Conductivity measurements show that concentrations of H3O+ and OH- in pure water are each only 1.07 x 10-7 mol/L of water at 25°C
That means in 10,000 L of water, less than 1 drop would be ionized

Standard notation represents concentration in moles per liter
Formula for ion or molecule in brackets [ ]
Ex. [H3O+] is concentration of hydronium in moles per liter
In water at 25°C [H3O+] = 1.0 x 10-7 M and [OH-] = 1.0 x 10-7 M

Mathematical product of [H3O+] and [OH-] remains constant in water and dilute aqueous solutions at constant temperature
This constant is called the ionization constant of water, Kw
Kw = [H3O+][OH-]
Kw = (1.0 x 10-7 M)(1.0 x 10-7 M) = 1.0 x 10-14 M2

Ionization of water increases as temperature increases
So, Kw increases as temp increases
At any given temp, Kw is always a constant value

Neutral, Acidic, and Basic Solutions
In pure water, [H3O+] = [OH-]
It is neutral
Any solution where [H3O+] = [OH-] is neutral
Remember acids increase [H3O+] in aqueous solutions
Acidic solutions: [H3O+] > [OH-]
Basic solutions: [H3O+] < [OH-]

Calculating [H3O+] and [OH-]
Strong acids and bases are considered completely ionized or dissociated in weak aqueous solutions
1 mol NaOH yields 1 mol OH- in aqueous solution
So 1.0 x 10-2 M NaOH has [OH-] of 1.0 x 10-2 M

Is greater than 1.0 x 10-7 M
Solution is basic
b/c Kw is relatively constant at 1.0 x 10-14 M2 at ordinary room temps, concentration of either ion can be calculated if concentration of 1 of them is known

Now consider a 1.0 x 10-4 M H2SO4 solution
Remember H2SO4 is diprotic
Acidic or basic?

Calculate [OH-]

The hydroxide ion concentration of an aqueous solution is 6.4 x 10-5 M. What is the hydronium concentration? (No, it’s not in your sheet)
1.6 x 10-10 M

Calculate the H3O+ and OH- concentrations in a 7.50 x 10-4 M solution of HNO3, a strong acid. (once again, NOT in your sheet ;-)
[H3O+] = 7.50 x 10-4 M
[OH-] = 1.33 x 10-11 M

Calculate [H3O+] and [OH-] of the following solutions. They are all strong acids/bases
a. 0.05 M sodium hydroxide
b. 0.0025 M sulfuric acid
c. 0.013 M lithium hydroxide
d. 0.150 M nitric acid
e. 0.0200 M calcium hydroxide
f. 0.390 M perchloric acid

a. 0.05 M sodium hydroxide
[OH-] = 0.05 M
[H3O+] = 2 x 10-13 M
b. 0.0025 M sulfuric acid
[OH-] = 2.0 x 10-12 M
[H3O+] = 5.0 x 10-3 M

c. 0.013 M lithium hydroxide
[OH-] = 0.013 M
[H3O+] = 7.7 x 10-13 M
d. 0.150 M nitric acid
[OH-] = 6.67 x 10-14 M
[H3O+] = 0.150 M

e. 0.0200 M calcium hydroxide
[OH-] = 0.0400 M
[H3O+] = 2.5 x 10-13 M
f. 0.390 M perchloric acid
[OH-] = 2.56 x 10-14 M
[H3O+] = 0.390 M

The pH Scale
pH stands for French words “pourvoirhydrogène” meaning hydrogen power
pH the negative of the common logarithm of the hydronium ion concentration, [H3O+]
pH = -log [H3O+]

Common logarithm of a number is the power to which 10 must be raised to equal the number
Neutral solution at room temp has [H3O+] of 1 x 10-7 M
Logarithm of 1 x 10-7 is -7.0

pOH the negative of the common logarithm of the hydroxide ion concentration [OH-]
pOH = -log [OH-]
Remember Kw = [H3O+][OH-] = 1.0 x 10-14
Negative log = 14.0
So pH + pOH = 14.0 at room temp

pH of acidic solution at 25°C with [H3O+] of 1 x 10-6 M is 6.0
pH is less than 7

Calculations Involving pH
If either [H3O+] or pH is known the other can be calculated
Significant figures involving pH must be handled carefully
b/c pH represents a log, the number to the left of the decimal only locates the decimal point
Not included when counting significant figures

There must be as many sig figs to the right of the decimal as there are in the number whose log was found
Ex. [H3O+] = 1 x 10-7
One sig fig
So, pH or –log must have one digit to the right of the decimal point
So, pH = 7.0

Calculating pH from [H3O+]
What is the pH of a 1.0 x 10-3 M NaOH solution?

1. Analyze
Given:
Identity and concentration of solution (1.0 x 10-3 M NaOH)
Unknown:
pH of solution

2. Plan
[base]  [OH-]  [H3O+]  pH

3. Compute

What is the pH of a solution with the following hydroxide ion concentrations?
a. 1 x 10-5 M
9.0
b. 2 x 10-8 M
6.7
c. 2.90 x 10-11 M
3.46

Using a Calculator to Calculate pH from [H3O+]
Some problems involve [H3O+] not equal to integral powers of 10
These require a calculator
Ex. What is the pH of a solution if the [H3O+] is 3.4 x 10-5 M?

What is the pH of a solution if the [H3O+] is 3.4 x 10-5 M?
The only difference is that you will determine the log of 3.4 x 10-5 M using a calculator
pH = -log [H3O+]
= -log (3.4 x 10-5)
= 4.47

(-)
Log
(
3.4
Shift
10x
(-)
5
)
=

What are the pOH values of solutions with the following [H3O+]?
a. 2.51 x 10-13 M
1.40
b. 4.3 x 10-3 M
11.6
c. 9.1 x 10-6 M
8.96
d. 0.070 M
12.8

A solution is prepared by dissolving 3.50 g of sodium hydroxide in water and adding water until the total volume of the solution is 2.50 L. What are [OH-] and [H3O+]?
[OH-] = 0.0350 M
[H3O+] = 2.86 x 10-13 M

Calculating [H3O+] and [OH-] from pH
You already know
pH = -log [H3O+]
Base of common logs is 10
So antilog of common log is 10 raised to that number
Log[H3O+] = antilog (-pH)
[H3O+] = 10-pH

Determine the [H3O+] of an aqueous solution with pH of 4.0
pH = -log [H3O+]
Log [H3O+] = -pH
[H3O+] = antilog (-pH)
[H3O+] = 1 x 10-pH
[H3O+] = 1 x 10-4 M

What are the pOH and [OH-] of a solution with a pH of 8.92?
pOH = 5.08
[OH-] = 8.3 x 10-6 M

A solution of NH3 has a pH of 11.00. Calculate [H3O+] and [OH-].
[H3O+] = 1.0 x 10-11 M
[OH-] = 1.0 x 10-3 M

Section 2
Determining pH and Titrations

Indicators and pH meters
Acid-base indicators compounds whose colors are sensitive to pH
They change color b/c they’re either weak acids or weak bases
In solution, equilibrium of indicator that is weak acid can be represented by
In- is the symbol of the anion part of the indicator

Colors that indicator displays result from the fact that HIn and In- are different colors

In acidic solutions, any In- ions that are present act as Brønsted-Lowry bases and accept protons from acid
Indicator is then present in mostly nonionized form, Hin
Indicator will have acid-indicating color (red)

In basic solutions, OH- ions from base combine with H+ made by indicator
Indicator molecules ionize to offset loss of H+ ions
Indicator present mostly in anion form, In-
Solution now shows base-indicating color, blue

Indicators come in different colors
Exact pH range also varies
Transition interval pH range over which an indicator changes color

Indicators that change color at pH lower than 7 (methyl orange) are stronger acids than other types of indicators
They ionize more than other indicators

The In- anions made by these are weaker Bronsted bases and are less likely to accept protons from any acid being tested
So they don’t shift to their nonionized (HIn) form unless concentration of H+ fairly high
Color transition happens at lower pH

In contrast, indicators that go through transition at higher pH (phenolphthalein) are weaker acids

Universal indicators made by mixing several different indicators
Paper soaked in universal indicator  pH paper

pH meter determines the pH of a solution by measuring voltage between two electrodes that are placed in solution
Voltage changes as hydronium ion concentration in solution changes

Titration
Neutralization reactions happen between acids and bases
OH- gets a proton from H3O+ forming 2 H2O

This shows that 1 mol hydronium (19.0g) and 1 mol hydroxide (17.0g) are chemically equal masses
They combine in 1:1 ratio
Neutralization occurs when hydronium and hydroxide ions are supplied in equal numbers by reactants

1 L of 0.10 M HCl contains 0.1 mol H3O+
Suppose 0.10 mol solid NaOH is added to 1 L of 0.10 M HCl
NaOH dissolves and supplies 0.10 mol OH-
HCl and NaOH are in chemically equal amounts
H3O+ and OH- combine until product [H3O+] [OH-] returns to 1.0 x 10-14 M2

Because acids and bases react, progressive addition of acid to base (or base to acid) can be used to compare concentrations of acid and base
Titration the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration

Equivalence Point
Equivalence point the point at which two solution used in a titration are present in chemically equal amounts
Indicators and pH meters used to determine the equivalence point
pH meter shows large voltage change at EP
End point  the point in a titration at which an indicator changes color

Some indicators, like litmus, change color at about pH 7
BUT the color change interval for litmus is wide, pH 5.5-8.0
Makes it difficult to determine accurate pH
Bromothymol blue is better
pH 6.0-7.6
Good for strong-acid/strong-base titrations

Indicators that change at pH lower than 7 useful to determine EP of strong-acid/weak-base titrations
Methyl orange, ex.
EP of strong-acid/weak-base titration is acidic b/c salt formed is a weak acid
So, salt solution has pH lower than 7

Indicators changing color at pH higher than 7 good for weak-acid/strong-base titrations
Phenolphthalein, ex.
These reactions make salt solutions pH higher than 7
This is b/c salt formed is weak base

As base is added, pH changes from low value to high one
Change in pH slow at first, then quickly through EP
Then slows down as solution becomes more acidic

Molarity and Titration
If the concentration of 1 solution is known, the concentration of the other solution in a titration can be calculated
Standard solution solution that contains the precisely known concentration of a solute
Often called “known solution”

To be certain of concentration of known solution, that solution must be compared with solution of a primary standard
Primary standard a highly purified solid compound used to check the concentration of the known solution in a titration

Known solution prepared first
Volume adjusted to give roughly the desired concentration
Concentration then determined more precisely by titrating it with a measured quantity of primary standard

Suppose 20.0 mL of 5.0 x 10-3 M NaOH required to reach end point in titration of 10.0 mL of HCl of unknown concentration
How can this data be used to determine molarity of acidic solution?

Start with balanced neutralization reaction equation
From equation, determine chemically equivalent amounts of HCl and NaOH

Calculate number of moles of NaOH used in titration
b/c 1 mol NaOH needed to neutralize 1 mol HCl, amount of HCl in titration must be 1.0 x 10-4 mol

This amount of acid must be in 10.0 mL of HCl used for titration
Now calculate molarity

Procedure
Start with balanced equation for neutralization reaction, and determine chemically equivalent amounts of acid and base.
Determine number of moles of acid/base from known solution used in titration.
Determine moles of solute of unknown.
Determine molarity of unknown solution.

Sample Problem
In a titration, 27.4 mL of 0.0154 M Ba(OH)2 is added to a 20.0 mL sample of HCl solution of unknown concentration. What is the molarity of the acid solution?

1. Analyze
Given:
Volume and concentration of known solution
27.4 mL of 0.0154 M Ba(OH)2
Volume of unknown HCl solution
20.0 mL
Unknown:
Molarity of acid solution

2. Plan
1. Balanced neutralization equation  chemically equivalent amounts
Ba(OH)2 + 2HCl  BaCl2 + 2H2O
1 mol 2 mol 1 mol 2 mol
Mole ratio is 1 mol Ba(OH)2 for every 2 mol HCl

2. Volume known basic solution used (mL)  amount of base used (mol)

3. Mole ratio, moles of base used  moles of acid used from unknown solution

4. Volume of unknown, moles of solute in unknown  molarity of unknown

A student titrates a 20.00 mL sample of a solution of HBr with unknown molarity. The titration requires 20.05 mL of a 0.1819 M solution of NaOH. What is the molarity of the HBr solution?
0.1824 M HBr

Vinegar can be assayed to determine its acetic acid content. Determine the molarity of acetic acid in a 15.00 mL sample of vinegar that requires 22.70 mL of a 0.550 M solution of NaOH to reach the equivalence point.
0.832 M

A 20.00 mL sample of a solution of Sr(OH)2 is titrated to the equivalence point with 43.03 mL of 0.1159 M HCl. What is the molarity of the Sr(OH)2 solution?
0.1247 M

A 35.00 mL sample of ammonia solution is titrated to the equivalence point with 54.95 mL of a 0.400 mL sulfuric acid solution. What is the molarity of the ammonia solution?
1.26 M

http://www.vinstan.com	332
http://vinstan.protonator.com	10
http://stfxsciencelab.blogspot.com	2
https://bb1011kdg.ua.ac.be	1

Chemistry Chapter 16 acid-base titration and ph

by Renee Ariesen on Jul 06, 2010

Accessibility

Categories

Tags

Upload Details

Usage Rights

4 Embeds 345

Statistics

Chemistry Chapter 16 acid-base titration and ph — Presentation Transcript