Chapter 16, Problem 16.21QP

Aqueous Solutions of Acids, Bases, and Salts

1. a For each of the following salts, write the reaction that occurs when it dissociates in water: NaCl(s), NaCN(s), KClO₂(s), NH₄NO₃(s), KBr(aq), and NaF(s).
2. b Consider each of the reactions that you wrote above, and identify the aqueous ions that could be proton donors (acids) or proton acceptors (bases). Briefly explain how you decided which ions to choose.
3. c For each of the acids and bases that you identified in pan b, write the chemical reaction it can undergo in aqueous solution (its reaction with water).
4. d Are there any reactions that you have written above that you anticipate will occur to such an extent that the pH of the solution will be affected? As pan of your answer, be sure to explain how you decided.
5. e Assume that in each case above, 0.01 mol of the salt was dissolved in enough water at 25°C to make 1.0 L of solution. In each case what additional information would you need in order to calculate the pH? If there are cases where no additional information is required, be sure to state that as well.
6. f Say you take 0.01 mol of NH₄CN and dissolve it in enough water at 25°C to make 1.0 L of solution. Using chemical reactions and words, explain how you would go about determining what effect this salt will have on the pH of the solution. Be sure to list any additional information you would need to arrive at an answer.

Interpretation Introduction

Interpretation:

The reaction of the given salts when it dissociates in water has to be written.

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

• Acidic-(pH will be less than seven)
• Basic -(pH will be more than seven)
• Neutral -(pH will be equal to seven)

To Write: The reaction of the given salts when it dissociates in water

Answer to Problem 16.21QP

The reaction of salt $\text{NaCl}(s)$ is $\text{NaCl}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + Cl}}^{-}(aq)$

The reaction of salt $\text{NaCN}(s)$ is $\text{NaCN}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + CN}}^{-}(aq)$

The reaction of salt ${\text{KClO}}_{\text{2}}(s)$ is ${\text{KClO}}_{\text{2}}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + ClO}}_{\text{2}}{}^{-}(aq)$

The reaction of salt ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)$ is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{+}(aq){\text{ + NO}}_{\text{3}}{}^{-}(aq)$

The reaction of salt $\text{KBr}(s)$ is $\text{KBr}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + Br}}^{-}(aq)$

The reaction of salt $\text{NaF}(s)$ is $\text{NaF}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + F}}^{-}(aq)$

Explanation of Solution

The reaction of the salt $\text{NaCl}(s)$ when it dissociates in water is:

$\text{NaCl}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + Cl}}^{-}(aq)$

The reaction of the salt $\text{NaCN}(s)$ when it dissociates in water is:

$\text{NaCN}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + CN}}^{-}(aq)$

The reaction of the salt ${\text{KClO}}_{\text{2}}(s)$ when it dissociates in water is:

${\text{KClO}}_{\text{2}}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + ClO}}_{\text{2}}{}^{-}(aq)$

The reaction of the salt ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)$ when it dissociates in water is:

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{+}(aq){\text{ + NO}}_{\text{3}}{}^{-}(aq)$

The reaction of the salt $\text{KBr}(s)$ when it dissociates in water is:

$\text{KBr}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + Br}}^{-}(aq)$

The reaction of the salt $\text{NaF}(s)$ when it dissociates in water is:

$\text{NaF}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + F}}^{-}(aq)$

Conclusion

The reaction of salt $\text{NaCl}(s)$ is written as $\text{NaCl}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + Cl}}^{-}(aq)$

The reaction of salt $\text{NaCN}(s)$ is written as $\text{NaCN}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + CN}}^{-}(aq)$

The reaction of salt ${\text{KClO}}_{\text{2}}(s)$ is written as ${\text{KClO}}_{\text{2}}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + ClO}}_{\text{2}}{}^{-}(aq)$

The reaction of salt ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)$ is written as ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(s)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{+}(aq){\text{ + NO}}_{\text{3}}{}^{-}(aq)$

The reaction of salt $\text{KBr}(s)$ is written as $\text{KBr}(s)\stackrel{}{\to }{\text{K}}^{+}(aq){\text{ + Br}}^{-}(aq)$

The reaction of salt $\text{NaF}(s)$ is written as $\text{NaF}(s)\stackrel{}{\to }{\text{Na}}^{+}(aq){\text{ + F}}^{-}(aq)$

Interpretation Introduction

Interpretation:

From the reactions given in part (a), the aqueous ions which are proton donors (acids) and proton acceptors (bases) has to be identified.

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

• Acidic-(pH will be less than seven)
• Basic -(pH will be more than seven)
• Neutral -(pH will be equal to seven)

To Identify: The proton donors (acids) and proton acceptors (bases) from part (a)

Answer to Problem 16.21QP

The ion which is a proton donor (acid) is ${\text{NH}}_{\text{4}}{}^{+}$

The ions which are a proton acceptors (bases) are ${\text{CN}}^{-},{\text{ ClO}}_{\text{2}}{}^{-}{\text{ and F}}^{-}$

Explanation of Solution

Ions as Proton donors:

The ion to be a proton donor must possess hydrogen and must be cation.

Thus, the only cation ion which possess hydrogen is ${\text{NH}}_{\text{4}}{}^{+}$

Hence, the ion which acts as proton donor (acid) is ${\text{NH}}_{\text{4}}{}^{+}$

Ions as Proton acceptors:

The proton acceptor must be an anion and are conjugate bases of weak acids.

The possible proton acceptors from the reactions in part (a) are ${\text{CN}}^{-},{\text{ ClO}}_{\text{2}}{}^{-}{\text{ and F}}^{-}$

Conclusion

The ion which is a proton donor (acid) is identified as ${\text{NH}}_{\text{4}}{}^{+}$

The ions which are a proton acceptors (bases) are identified as ${\text{CN}}^{-},{\text{ ClO}}_{\text{2}}{}^{-}{\text{ and F}}^{-}$

Interpretation Introduction

Interpretation:

The chemical reaction in water for each of the acids and bases identified in part (b) has to be written.

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

• Acidic-(pH will be less than seven)
• Basic -(pH will be more than seven)
• Neutral -(pH will be equal to seven)

To Write: The chemical reaction in water for each of the acids and bases identified in part (b)

Answer to Problem 16.21QP

The chemical reaction in water for each of the acids and bases identified in part (b) are:

${\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) +H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + NH}}_{\text{3}}\text{(aq)}$

${\text{CN}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HCN(aq) + OH}}^{-}\text{(aq)}$

${\text{ClO}}_{\text{2}}{}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HClO}}_{\text{2}}{\text{(aq) + OH}}^{-}\text{(aq)}$

${\text{F}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HF(aq) + OH}}^{-}\text{(aq)}$

Explanation of Solution

Chemical reaction in water for the acids and bases:

The ion ${\text{NH}}_{\text{4}}{}^{+}$ which is a proton donor (acid) reacts with water as follows,

${\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) +H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + NH}}_{\text{3}}\text{(aq)}$

The ion ${\text{CN}}^{-}$ which is a proton acceptor (base) reacts with water as follows,

${\text{CN}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HCN(aq) + OH}}^{-}\text{(aq)}$

The ion ${\text{ClO}}_{\text{2}}{}^{-}$ which is a proton acceptor (base) reacts with water as follows,

${\text{ClO}}_{\text{2}}{}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HClO}}_{\text{2}}{\text{(aq) + OH}}^{-}\text{(aq)}$

The ion ${\text{F}}^{-}$ which is a proton acceptor (base) reacts with water as follows,

${\text{F}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HF(aq) + OH}}^{-}\text{(aq)}$

Conclusion

The chemical reaction in water for each of the acids and bases identified in part (b) are:

${\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) +H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + NH}}_{\text{3}}\text{(aq)}$

${\text{CN}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HCN(aq) + OH}}^{-}\text{(aq)}$

${\text{ClO}}_{\text{2}}{}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HClO}}_{\text{2}}{\text{(aq) + OH}}^{-}\text{(aq)}$

${\text{F}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HF(aq) + OH}}^{-}\text{(aq)}$

Interpretation Introduction

Interpretation:

Does any of the reactions written in part (c) will occur to such an extent that the pH of the solution will be affected has to be explained

Concept Introduction:

Relationship between $K_{a}and{K}_b$

$\begin{array}{l}{\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}\\ \text{ or}\\ {\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\end{array}$

Where ${\text{K}}_{\text{w}}$ is the auto-ionization of water. $(1.0\times {10}^{-14})$

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To Explain: Does any of the reactions written in part (c) will occur to such an extent that the pH of the solution will be affected

Explanation of Solution

On seeing at the $\text{K}$ value for each of the above reaction in part (c), we can get an idea of the extent of these reactions.

We know that, ${\text{K}}_{\text{a}}$ is inversely proportional to  ${\text{K}}_b$ which means, when the $\text{K}$ value for an acid or base is smaller, the $\text{K}$ value for its conjugate acid or base will be higher.

Every acid will have a conjugate base.

For the acids with smaller ${\text{K}}_{\text{a}}$ value, its conjugate base undergoes hydrolysis to the greatest extent and have greatest effect on the pH of the solution whereas for the acids with larger ${\text{K}}_{\text{a}}$ value, its conjugate base undergoes hydrolysis to the lesser extent and have smaller effects on the pH of a solution.

Conclusion

Does any of the reactions written in part (c) will occur to such an extent that the pH of the solution will be affected was explained.

Interpretation Introduction

Interpretation:

On assuming that $0.01\text{ mol}$ of each of the salt is dissolved in enough of water to make $0.1\text{ L}$ of solution, the additional informations that are either needed or not needed in order to calculate the pH for each of the solution has to be explained.

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

• Acidic-(pH will be less than seven)
• Basic -(pH will be more than seven)
• Neutral -(pH will be equal to seven)

Explanation of Solution

For the solutions made from either ${\text{NaCN, KClO}}_{\text{2}},{\text{ NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{ or NaF}$ salt.

The ${\text{K}}_{\text{a}}$ value or ${\text{K}}_b$ for the corresponding acid or base is needed to calculate the pH.

For the solutions made from $\text{NaCl or KBr}$, no further information is needed to calculate the pH, because none of the ions hydrolyze, so the pH of the solution would be 7.00

Conclusion

The additional informations that are either needed or not needed in order to calculate the pH for each of the given solutions was explained

Interpretation Introduction

Interpretation:

The effect of salt ${\text{NH}}_{\text{4}}\text{CN}$ on determining the pH of the solution of same salt and the additional information needed to arrive at the answer has to be explained.

Concept Introduction:

Relationship between $K_{a}and{K}_b$

$\begin{array}{l}{\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}\\ \text{ or}\\ {\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\end{array}$

Where ${\text{K}}_{\text{w}}$ is the auto-ionization of water. $(1.0\times {10}^{-14})$

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To Explain: The effect of salt ${\text{NH}}_{\text{4}}\text{CN}$ on determining the pH of the solution of same salt and the additional information needed to arrive at the answer

Explanation of Solution

Given data:

A 0.01 mol of ${\text{NH}}_{\text{4}}\text{CN}$ is dissolved in enough of water to make $0.1\text{ L}$ of solution.

Effect of salt on pH calculation:

From the given data, we know the concentration of the solution would be $0.010\text{ M}$.

The soluble  ${\text{NH}}_{\text{4}}\text{CN}$ salt produce a solution consisting of $0.010{\text{ M NH}}_{\text{4}}{}^{\text{+}}$ ions and $0.010{\text{ M CN}}^{-}$

Both the ions undergoes hydrolysis as follows,

${\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) +H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + NH}}_{\text{3}}\text{(aq)}$

${\text{CN}}^{-}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\stackrel{}{\rightleftharpoons }{\text{ HCN(aq) + OH}}^{-}\text{(aq)}$

Since ${\text{NH}}_{\text{4}}{}^{\text{+}}$ would produce an acidic solution and ${\text{CN}}^{\text{-}}$ would produce a basic solution, the ions have opposite effects on the pH of the solution.

Therefore, the $\text{K}$ value for both the reactions are compared and the reaction having larger $\text{K}$ value will be dominating the pH.

Both of these reactions involve a conjugate acid or base, so it is necessary to compare the $\text{K}$ values for the acid or base.

The smaller the $\text{K}$ value for an acid or base, the larger the $\text{K}$ value for its conjugate.

we need the ${\text{K}}_{\text{b}}$ for ${\text{NH}}_{\text{3}}$ and the ${\text{K}}_a$ value for $\text{HCN}$. If ${\text{K}}_{\text{a}}{\text{ < K}}_{\text{b}}$ , the solution will be basic, and if ${\text{K}}_{\text{a}}{\text{ > K}}_{\text{b}}$, the solution will be acidic.

Conclusion

The effect of salt ${\text{NH}}_{\text{4}}\text{CN}$ on determining the pH of the solution of same salt and the additional information needed to arrive at the answer was explained.