Acid and bases differentiated assignment p h_answers
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Student A's statement that a strong acid has a low pH and a weak acid has a high pH is correct only if comparing acids of the same concentration and basicity. Student B's statement that a dibasic acid has a lower pH than a monobasic acid is only correct if the acids are equally strong and concentrated. The pH of a 0.01 M NaOH solution is calculated to be 12, and the pH of a 0.1 M HCl solution before titration with 0.1 M NaOH is calculated to be 1. After adding 5 cm3 and 24 cm3 of NaOH, the pH is calculated to be 1.18 and 2.69 respectively. The pH after 26 cm3 of
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![Answers to Differentiated Assignment – Acids and Bases : pH<br />1Student A wrote:<br />“A strong acid is one which has a low pH and a weak acid is one which has high pH.”<br />Student B wrote:<br />“A dibasic acid has a lower pH than a monobasic acid.”<br />Comment on the assumptions made by<br />Student A’s statement is correct only if the acids compared are of the same concentration and the same basicity.[2]<br />(also, “high pH” is not clearly defined as it could mean a number close to 7 or exceeding 7)<br />Student B’s statement is correct only if the two acids compared are equally strong and of the same concentration.[2]<br />2Given that pOH can be used to measure the concentration OH- ions in a solution and is given by the following formula:<br />pOH = -log[OH-]<br />The relationship between pH and pOH is given by the following equation:<br />pH + pOH = 14<br />a.Calculate the pOH of NaOH (aq) of concentration 0.01M<br />[OH-] = 0.01M = 1 X 10-2 mol/dm3 <br />pOH = - log [OH-] = - log (1 X 10 -2) <br />pOH = - (-2) <br />pOH = 2[1]<br />What is the pH of this solution?<br />pH = 14 – 2 = 12[1]<br />c*Calculate [H+] from your answer to (b).<br />pH = -log10[H+]<br />[H+]= 10-pH<br />[H+] = 10–12 mol/dm3[1]<br />325.0 cm3 of 0.100 M hydrochloric acid in a conical flask was titrated against 0.100 M of sodium hydroxide from a burette. The reaction can be represented by the following equation:<br />HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)<br />a.Calculate the pH of the acid solution before the titration<br />pH = -log[0.1] = 1[1]<br />b.The pH of the mixture in the conical flask was monitored during the titration <br />by a pH electrode. When 5.0 cm3 of sodium hydroxide solution was added to the acid, calculate<br />i.the amount of H+ remaining in the conical flask.<br />No. of moles of H+ before titration = 0.1 M x 0.025 dm3 = 0.0025<br />No. of moles of H+ reacted with NaOH = 0.1 M x 0.005dm3 = 0.0005<br />No. of moles of H+ remaining = 0.0025 – 0.0005 = 0.002[3]<br />ii.the new volume of mixture in the conical flask.<br />New volume of mixture = 0.025 + 0.005 = 0.030 dm3[1]<br />iii.the new concentration of H+ in the conical flask<br />[H+] after 5 cm3 of NaOH was added = 0.002/0.030 mol/dm3 =0.0667 M[1]<br />iv.the new pH of the mixture in the conical flask<br />pH = -log [0.0667] = 1.176 ≈ 1.18[1]<br />c*Following the method used in (b), or otherwise, calculate the pH of the <br />mixture in the conical flask after 24.0 cm3 of sodium hydroxide <br />has been added:<br />[H+] before titration = 0.1 M<br />No. of moles of H+ before titration = 0.1 M x 0.025 dm3 = 0.0025 [0.5]<br />No. of moles of H+ reacted with NaOH = 0.1 M x 0.024 dm3 = 0.0024 [0.5]<br />No. of moles of H+ remaining = 0.0025 – 0.0024 = 0.0001<br />New volume of mixture = 0.025 + 0.024 = 0.049 dm3 [0.5]<br />[H+] after 24 cm3 of NaOH was added = 0.0001/0.049 mol/dm3 =0.00204 M<br />pH = -log [0.00204] = 2.69 [0.5]<br />d.Would the pH of the mixture in the conical flask after 26.0 cm3 of sodium hydroxide has been added be closer to pH 7 or 11? Explain your choice.<br />It is closer to pH 11.<br />After 26 cm3 of NaOH is added it is past the end-point thus all acid is used up. [1]<br />Since NaOH is a strong alkali, an excess will show a high pH.[1]<br />e.Sketch a graph of the pH vs the volume of sodium hydroxide added using the <br />values you have calculated.<br />End point = 7<br />Volume of NaOH addedpH<br />Equivalence point <br />volume of alkali added<br />shape of graph [0.5m]<br />](https://image.slidesharecdn.com/acidandbasesdifferentiatedassignmentphanswers-111006032015-phpapp02/85/Acid-and-bases-differentiated-assignment-p-h_answers-1-320.jpg)
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Here are potential responses to the study questions:
Define the following terms:
- Ionization: The process by which an atom or molecule acquires a negative or positive charge by gaining or losing electrons.
- Buffer capacity: The ability of a solution to resist changes in pH upon the addition of an acid or base. It depends on the buffer composition and concentration.
- In-vivo: Occurring or taking place inside a living organism.
Considering a practical process, illustrate the procedural significance of buffer systems in moderation of the reactions of a solution system:
Buffer systems are important in pharmaceutical formulations to maintain the pH within an optimal range for drug stability, solubility, and to minimize irritation upon administration.
11 buffer solutions in pharmacy
Here are potential responses to the study questions:
Define the following terms:
- Ionization: The process by which an atom or molecule acquires a negative or positive charge by gaining or losing electrons.
- Buffer capacity: The ability of a solution to resist changes in pH upon the addition of an acid or base. It depends on the buffer composition and concentration.
- In-vivo: Occurring inside a living organism.
Considering a practical process, illustrate the procedural significance of buffer systems in moderation of the reactions of a solution system:
Buffer systems are important in pharmaceutical formulations to maintain the pH within an optimal range for drug stability, solubility, and to prevent irritation upon administration. They moderate changes
Acid-Base-Equilibria.ppt
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Acids and bases p pt
This document defines acids and bases according to the Arrhenius model and discusses their properties in aqueous solutions. It also explains acid-base titrations for determining concentrations and defines pH. Buffers are formed from a weak acid and its conjugate base or weak base and conjugate acid and do not experience large pH changes upon addition of acids or bases.
Ch15 z5e aq. equil
This document provides an overview of applying equilibrium concepts to acid-base chemistry. It discusses the common ion effect, buffered solutions, calculating pH of buffers using the Henderson-Hasselbach equation, effects of adding strong acids/bases to solutions, buffer capacity, and titration curves. Key points include:
- The common ion effect lowers the dissociation of a weak acid/base when a salt containing its ion is added
- Buffered solutions resist pH changes due to a mixture of weak acid/base and its salt
- The H-H equation can be used to calculate pH of buffer solutions
- Adding a strong base/acid may react with and reduce the concentration of the conjugate acid/base of a buffer
The PH Scale
The document discusses pH and the pH scale. It defines pH as the negative logarithm of the molar concentration of hydrogen ions and explains that pH values below 7 indicate increasing acidity while values above 7 indicate increasing basicity. It provides examples of calculating pH from given hydrogen ion concentrations and vice versa. It also discusses buffer solutions and how they resist changes in pH when small amounts of acid or base are added.
Chapter_14_-_Acids_and_Bases.ppt
This document provides an overview of acids and bases, including:
- Definitions of acids and bases according to Arrhenius, Brønsted-Lowry, and Lewis theories.
- Characteristics of strong vs. weak acids and bases, and how their ionization depends on concentration and acid/base strength.
- The concept of conjugate acid-base pairs and how acid/base strength relates to the strength of their conjugates.
- How to determine pH, pOH, and percent ionization for strong/weak acids and bases using ionization constants and ICE tables.
- How pH indicates whether a solution is acidic, basic, or neutral.
Neutralization titrations
* HCl is a strong acid and will titrate first
* Its equivalence point was at 35.00 mL of NaOH
* NaOH concentration is 0.100 M
* Moles of NaOH used = Volume x Concentration
= 0.03500 L x 0.100 mol/L = 0.003500 mol
* Moles of HCl = Moles of NaOH used = 0.003500 mol
* H3PO4 is a weak acid and will titrate second
* Its equivalence point was at 50.00 mL of NaOH
* Additional NaOH used = 50.00 mL - 35.00 mL = 15.00 mL
* Moles of additional Na
Neutralizationtitrations 140617143109-phpapp01
* HCl is a strong acid and will titrate first
* Its equivalence point was at 35.00 mL of NaOH
* NaOH concentration is 0.100 M
* Moles of NaOH used = Volume x Concentration
= 0.03500 L x 0.100 mol/L = 0.003500 mol
* Moles of HCl = Moles of NaOH used = 0.003500 mol
* H3PO4 is a weak acid and will titrate second
* Its equivalence point was at 50.00 mL of NaOH
* Additional NaOH used = 50.00 mL - 35.00 mL = 15.00 mL
* Moles of additional Na
Chapter 1 acid and bases sesi 2
The document discusses acids and bases according to various theories including Arrhenius and Bronsted-Lowry. It defines acids as hydrogen ion donors and bases as hydrogen ion acceptors. Acids are classified as strong or weak based on their degree of ionization in water. Buffer solutions are introduced as mixtures that minimize pH changes from the addition of small amounts of acid or base. Common examples of acidic and alkaline buffer solutions are provided.
Acid-Base Equilibria.ppt
This document discusses acid-base equilibria, including definitions of acids and bases according to Bronsted-Lowry theory. It covers water's behavior as both an acid and a base, pH calculations, strong and weak acids/bases, and polyprotic acids. Key points addressed are: the autoionization of water, conjugate acid-base pairs, the relationship between Ka and Kb, and using Ka or Kb values to determine pH or calculate concentrations.
Chapter 16.1 and 2 : Acid-Base Titrations and pH
The document discusses acid-base titrations and pH. It defines pH as the negative logarithm of hydronium ion concentration and explains how to calculate pH from [H3O+] and vice versa. It also describes how to perform and calculate molarity from a titration experiment by determining moles of acid/base reacted using volume and molarity of the titrant. Sample titration problems are worked through demonstrating these concepts.
Apchemunit15presentation 120311192458-phpapp01
1. This unit covers acid-base chemistry including solutions containing a common ion, buffered solutions, and titrations.
2. Buffered solutions contain relatively large concentrations of a weak acid and its conjugate base or a weak base and its conjugate acid. They resist changes in pH when small amounts of acid or base are added.
3. Titration curves are generated by plotting pH versus volume of titrant added. The pH changes slowly until near the equivalence point, where a small change in titrant produces a large pH change as the ratio of [H+]/[OH-] changes dramatically.
Ap chem unit 15 presentation
Here are the ratios [HA]/[A-] required for each system to yield a pH of 4.30:
Acetic acid (pKa = 4.76): [HA]/[A-] = 1
Benzoic acid (pKa = 4.19): [HA]/[A-] = 1
Phthalic acid (pKa = 2.89): [HA]/[A-] is not possible since the pKa is too low.
The optimal system is acetic acid/sodium acetate since it has a pKa closest to the desired pH of 4.30. This system allows a ratio of [HA]/[A-] = 1, which provides maximum buffering
Acid,Bases And Salt(s) (ScienCe)
This document discusses acids, bases, and salts. It defines acids as sour tasting substances that increase the hydronium ion concentration in water. Bases are defined as bitter tasting and slippery substances that increase the hydroxide ion concentration in water. Strong acids and bases are completely ionized in water while weak acids and bases are only partially ionized. The document also discusses pH, the logarithmic scale used to measure acidity and alkalinity, and how pH and pOH are related through their sum being equal to 14.
acid-base-equilebria.ppt
This document discusses acid-base concepts including:
- Arrhenius and Bronsted-Lowry definitions of acids and bases
- Conjugate acid-base pairs and the Lewis definitions
- Acid-base reactions and the Brønsted-Lowry definitions of acids donating H+ and bases accepting H+
- Acid ionization constants (Ka) and base ionization constants (Kb) as well as the water autoionization constant (Kw) and how it relates to pH
- Calculations involving pH, pOH, [H+], Ka, Kb, and the Henderson-Hasselbalch equation for buffer solutions
Chapter 5&6
analytical Chemistry just the basic one. consists of compleximetric titration and the principle titration
Mod2 3. Acid Base Equilibria.pptx
This document outlines the key objectives and concepts around acid-base equilibria, including:
- Defining strong and weak acids/bases using Bronsted-Lowry theory and discussing conjugate acid-base pairs
- Explaining the pH scale and relating pH, pOH, Ka, and pKa values
- Describing how to calculate the pH of strong acids/bases from their concentrations and vice versa
- Discussing how weak acids only partially dissociate in solution according to their acid dissociation constant (Ka)
- Demonstrating calculations for finding the pH of a solution of a weak acid using its Ka value
Acids-Bases-pH-PPT-2018.ppt.organic-chemistry
This document discusses the chemistry of acids and bases. It defines acids as substances that produce hydrogen ions (H+) or hydronium ions (H3O+) in water, and bases as substances that produce hydroxide ions (OH-). It provides examples of common acids and bases and discusses their properties. Key topics covered include the Arrhenius, Brønsted-Lowry, and Lewis definitions of acids and bases; acid-base reactions; pH and pOH scales; strong and weak acids and bases; and acid-base equilibria.
Similar to Acid and bases differentiated assignment p h_answers (20)
CHM025-Topic-V_Acid-Base-Titrimetry_231215_154125.pdf
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Kinetic questions about doubling the volume of reactants
Graph A shows the volume of hydrogen gas produced over time from the reaction of 10g of magnesium with 400cm3 of 1M HCl.
Graph B would show a lower initial rate and lower maximum volume, as only 5g of magnesium is used with the same volume of 2M HCl.
Graph C would show a higher initial rate and reach a higher maximum volume than Graph A, as 50cm3 of a more concentrated hydrogen peroxide solution is used with the same amount of catalyst.
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The document contains calculations to determine the ages of various chemists based on chemical reactions and analyses provided. It is determined that:
- Dabao is 64 years old based on Graham's law of diffusion calculations.
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pH - Understanding titration curve
This document summarizes different types of acid-base titration curves:
1) A strong acid with a strong base shows a steep plunge in pH from 10.5 to 3.5 at the equivalence point of 25 mL, indicating the reaction of all H+ and OH- ions.
2) A strong acid with a weak base resembles the strong acid-strong base curve before the equivalence point and the weak acid-strong base curve after. The endpoint pH is slightly acidic due to the acidic ammonium chloride product.
3) A strong base with a weak acid shows a basic endpoint due to the alkaline sodium acetate product. The curve is less steep than the strong acid reactions due to the weak acid-base interaction.
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This document describes the process of preparing salts through titration. It explains that titration allows the neutralization of an acid and base to be carried out exactly, producing a soluble salt. It provides the example of preparing sodium chloride through titrating hydrochloric acid with sodium hydroxide. The steps involve adding an indicator, titrating the acid with the base until the endpoint is reached, evaporating the solution to leave behind salt crystals, and filtering and drying the crystals. It asks how ammonium nitrate could be prepared using this method.
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Acid and bases differentiated assignment p h_answers
- 1. Answers to Differentiated Assignment – Acids and Bases : pH<br />1Student A wrote:<br />“A strong acid is one which has a low pH and a weak acid is one which has high pH.”<br />Student B wrote:<br />“A dibasic acid has a lower pH than a monobasic acid.”<br />Comment on the assumptions made by<br />Student A’s statement is correct only if the acids compared are of the same concentration and the same basicity.[2]<br />(also, “high pH” is not clearly defined as it could mean a number close to 7 or exceeding 7)<br />Student B’s statement is correct only if the two acids compared are equally strong and of the same concentration.[2]<br />2Given that pOH can be used to measure the concentration OH- ions in a solution and is given by the following formula:<br />pOH = -log[OH-]<br />The relationship between pH and pOH is given by the following equation:<br />pH + pOH = 14<br />a.Calculate the pOH of NaOH (aq) of concentration 0.01M<br />[OH-] = 0.01M = 1 X 10-2 mol/dm3 <br />pOH = - log [OH-] = - log (1 X 10 -2) <br />pOH = - (-2) <br />pOH = 2[1]<br />What is the pH of this solution?<br />pH = 14 – 2 = 12[1]<br />c*Calculate [H+] from your answer to (b).<br />pH = -log10[H+]<br />[H+]= 10-pH<br />[H+] = 10–12 mol/dm3[1]<br />325.0 cm3 of 0.100 M hydrochloric acid in a conical flask was titrated against 0.100 M of sodium hydroxide from a burette. The reaction can be represented by the following equation:<br />HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)<br />a.Calculate the pH of the acid solution before the titration<br />pH = -log[0.1] = 1[1]<br />b.The pH of the mixture in the conical flask was monitored during the titration <br />by a pH electrode. When 5.0 cm3 of sodium hydroxide solution was added to the acid, calculate<br />i.the amount of H+ remaining in the conical flask.<br />No. of moles of H+ before titration = 0.1 M x 0.025 dm3 = 0.0025<br />No. of moles of H+ reacted with NaOH = 0.1 M x 0.005dm3 = 0.0005<br />No. of moles of H+ remaining = 0.0025 – 0.0005 = 0.002[3]<br />ii.the new volume of mixture in the conical flask.<br />New volume of mixture = 0.025 + 0.005 = 0.030 dm3[1]<br />iii.the new concentration of H+ in the conical flask<br />[H+] after 5 cm3 of NaOH was added = 0.002/0.030 mol/dm3 =0.0667 M[1]<br />iv.the new pH of the mixture in the conical flask<br />pH = -log [0.0667] = 1.176 ≈ 1.18[1]<br />c*Following the method used in (b), or otherwise, calculate the pH of the <br />mixture in the conical flask after 24.0 cm3 of sodium hydroxide <br />has been added:<br />[H+] before titration = 0.1 M<br />No. of moles of H+ before titration = 0.1 M x 0.025 dm3 = 0.0025 [0.5]<br />No. of moles of H+ reacted with NaOH = 0.1 M x 0.024 dm3 = 0.0024 [0.5]<br />No. of moles of H+ remaining = 0.0025 – 0.0024 = 0.0001<br />New volume of mixture = 0.025 + 0.024 = 0.049 dm3 [0.5]<br />[H+] after 24 cm3 of NaOH was added = 0.0001/0.049 mol/dm3 =0.00204 M<br />pH = -log [0.00204] = 2.69 [0.5]<br />d.Would the pH of the mixture in the conical flask after 26.0 cm3 of sodium hydroxide has been added be closer to pH 7 or 11? Explain your choice.<br />It is closer to pH 11.<br />After 26 cm3 of NaOH is added it is past the end-point thus all acid is used up. [1]<br />Since NaOH is a strong alkali, an excess will show a high pH.[1]<br />e.Sketch a graph of the pH vs the volume of sodium hydroxide added using the <br />values you have calculated.<br />End point = 7<br />Volume of NaOH addedpH<br />Equivalence point <br />volume of alkali added<br />shape of graph [0.5m]<br />