Chapter 16, Problem 16.22QP

The pH of Mixtures of Acid, Base, and Salt Solutions

1. a When 0.10 mol of the ionic solid NaX, where X is an unknown anion, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 9.12. When 0.10 mol of the ionic solid ACl, where A is an unknown cation, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 7.00. What would be the pH of 1.0 L of solution that contained 0.10 mol of AX? Be sure to document how you arrived at your answer.
2. b In the AX solution prepared above, is there any OH⁻ present? If so, compare the [OH⁻] in the solution to the [H₃O⁺].
3. c From the information presented in part a, calculate K_b for the X⁻(aq) anion and K_a for the conjugate acid of X⁻(aq).
4. d To 1.0 L of solution that contains 0.10 mol of AX, you add 0.025 mol of HCl. How will the pH of this solution compare to that of the solution that contained only NaX? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer.
5. e Another 1.0 L sample of solution is prepared by mixing 0.10 mol of AX and 0.10 mol of HCl. The pH of the resulting solution is found to be 3.12. Explain why the pH of this solution is 3.12.
6. f Finally, consider a different 1.0-L sample of solution that contains 0.10 mol of AX and 0.1 mol of NaOH. The pH of this solution is found to be 13.00. Explain why the pH of this solution is 13.00.
7. g Some students mistakenly think that a solution that contains 0.10 mol of AX and 0.10 mol of HCl should have a pH of 1.00. Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.

Interpretation Introduction

Interpretation:

The steps for the calculation of pH of $1.0\text{ L}$ solution of $0.1\text{ mol}$ of unknown $\text{AX}$ 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)

To Explain: The steps for the calculation of pH of $1.0\text{ L}$ solution of $0.1\text{ mol}$ of unknown $\text{AX}$

Answer to Problem 16.22QP

The pH of the solution prepared from the ionic compound $\text{AX}$ will be 9.12

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{NaX}$, in water. The pH of this solution is 9.12 and $\text{X}$ is an unknown anion.

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{ACl}$, in water. The pH of this solution is 7.00 and $\text{A}$ is an unknown cation.

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$, in water

Calculation of pH of $\text{AX}$solution:

The ionic compound $\text{NaX}$ dissociates into ${\text{Na}}^{+}$ and ${\text{X}}^{-}$ ion. ${\text{X}}^{-}$ is the conjugate base of a weak acid  whereas ${\text{Na}}^{+}$ is a neutral ion. The ${\text{X}}^{-}$ ion undergoes hydrolysis in water and gives a basic solution with a pH of 9.12

The ionic compound $\text{ACl}$ dissociates into ${\text{A}}^{+}$ and ${\text{Cl}}^{-}$ ion. Since the pH of the solution is 7.00, it implies neutral solution. The ${\text{Cl}}^{-}$ ion is a neutral and ${\text{A}}^{+}$ ion is also a neutral because the given solution is neutral.

The ionic compound $\text{AX}$ dissociates into ${\text{A}}^{+}$ and ${\text{X}}^{-}$ ions. Since, ${\text{A}}^{+}$ ions are neutral, ${\text{X}}^{-}$ ions hydrolyze in water and produce a solution with pH of 9.12, which is the same as for the solution of $\text{NaX}$

Conclusion

The pH of the solution prepared from the ionic compound $\text{AX}$ will be 9.12

Interpretation Introduction

Interpretation:

Does the $\text{AX}$ solution prepared in part (a) has ${\text{OH}}^{-}$ ions has to be explained and if present, ${\text{[OH}}^{-}]$ has to be compared with the ${\text{[H}}_{\text{3}}{\text{O}}^{+}]$

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)

Answer to Problem 16.22QP

The ${\text{OH}}^{-}$ ions are present in the $\text{AX}$ solution.

On comparison of concentration, ${\text{[OH}}^{-}]{\text{ > [H}}_{\text{3}}{\text{O}}^{+}]$

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$, in water

This solution is found be basic with pH of 9.12 (from part (a))

${\text{OH}}^{-}$ ion and ${\text{[OH}}^{-}]$:

Since the ${\text{X}}^{-}$ ion of the compound $\text{AX}$ undergoes hydrolysis, it gives ${\text{OH}}^{-}$ ions in solution. Therefore, the given solution of $\text{AX}$ contains ${\text{OH}}^{-}$ ions

Since the solution is a basic solution, the concentration of hydroxide ions will be greater than the concentration of hydronium ions.

Thus, ${\text{[OH}}^{-}]{\text{ > [H}}_{\text{3}}{\text{O}}^{+}]$

Conclusion

The ${\text{OH}}^{-}$ ions are present in the $\text{AX}$ solution.

On comparison of concentration, ${\text{[OH}}^{-}]{\text{ > [H}}_{\text{3}}{\text{O}}^{+}]$

Interpretation Introduction

Interpretation:

Using part (a) information, the ${\text{K}}_{\text{b}}$ for the ${\text{X}}^{-}$ anion and ${\text{K}}_a$ for the conjugate acid of ${\text{X}}^{-}$ ion has to be calculated.

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. $(1.0\times {10}^{-14})$

To Calculate: Using part (a) information, the ${\text{K}}_{\text{b}}$ for the ${\text{X}}^{-}$ anion and ${\text{K}}_a$ for the conjugate acid of ${\text{X}}^{-}$ ion

Answer to Problem 16.22QP

The ${\text{K}}_{\text{b}}$ for the ${\text{X}}^{-}$ anion is $1.7\times {10}^{-9}$

${\text{K}}_a$ for the conjugate acid of ${\text{X}}^{-}$ ion is $5.8\times {10}^{-6}$

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$, in water

This solution is found be basic with pH of 9.12 (from part (a))

Calculation of ${\text{K}}_a$ and ${\text{K}}_{\text{b}}$:

A $1.0\text{ L}$ solution of $\text{AX}$ contains $0.1\text{ mol}$ $\text{AX}$.

Thus, $\text{[AX]}$= $0.1\text{ mol}$

The hydrolysis of ${\text{X}}^{-}$ is:

$\begin{array}{l}{\text{X}}^{\text{-}}(aq){\text{ + H}}_{\text{2}}\text{O}(l)\stackrel{}{\rightleftarrows }\text{ HX}(aq){\text{ + OH}}^{{}^{-}}(aq)\\ 0.10-xxx\end{array}$

The ${\text{[OH}}^{-}]$ is calculated using pH.

$\begin{array}{l}\text{pOH }\text{= 14}\text{.00 - pH}\\ \text{=14}\text{.00 - 9}\text{.12}\\ \text{=4}\text{.88}\end{array}$

Therefore,

$\begin{array}{l}x={\text{[OH}}^{\text{-}}\text{]}\\ ={10}^{-\text{pOH}}\\ ={10}^{-4.88}\\ =1.318\times {10}^{-5}\end{array}$

The ${\text{K}}_{\text{b}}$ is calculated as follows,

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{x^2}{0.10-x}\\ =\frac{{(1.318\times {10}^{-5})}^2}{0.10-1.318\times {10}^{-5}}\\ =1.7\times {10}^{-9}\end{array}$

The ${\text{K}}_a$ is calculated as follows,

$\begin{array}{l}{\text{ K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\\ =\frac{\text{1}\text{.00}\times {\text{10}}^{-14}}{\text{1}\text{.7}\times {\text{10}}^{-9}}\\ =5.8\times {\text{10}}^{-6}\end{array}$

Conclusion

The ${\text{K}}_{\text{b}}$ for the ${\text{X}}^{-}$ anion is calculated as $1.7\times {10}^{-9}$

${\text{K}}_a$ for the conjugate acid of ${\text{X}}^{-}$ ion is calculated as $5.8\times {10}^{-6}$

Interpretation Introduction

Interpretation:

The pH of the $1.0\text{ L}$ of $0.1\text{ mol}$ $\text{AX}$ solution containing $0.025\text{ mol}$ of $\text{HCl}$ has to be compared with the solution containing only $\text{NaX}$

To Compare: The pH of the $1.0\text{ L}$ of $0.1\text{ mol}$ $\text{AX}$ solution containing $0.025\text{ mol}$ of $\text{HCl}$ with the solution containing only $\text{NaX}$

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{NaX}$, in water. The pH of this solution is 9.12 and $\text{X}$ is an unknown anion.

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{ACl}$, in water. The pH of this solution is 7.00 and $\text{A}$ is an unknown cation.

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$, in water

This solution is basic and has a pH of 9.12

Comparison of pH:

On addition of $\text{HCl}$ to the solution of $\text{AX}$, ${\text{X}}^{-}$ is removed from the solution since it is replaced by equal moles of $\text{HX}$.

Since the $\text{HCl}$ added is not enough to completely react with all the ${\text{X}}^{-}$ present, the resulting solution will still be basic. But the pH of this resulting solution will be lower.

Since the solution of $\text{NaX}$ and the solution of $\text{AX}$ have the same conjugate base with the same concentration, the effect on both solutions is the same.

Conclusion

The pH of the $1.0\text{ L}$ of $0.1\text{ mol}$ $\text{AX}$ solution containing $0.025\text{ mol}$ of $\text{HCl}$ is compared with the solution containing only $\text{NaX}$

Interpretation Introduction

Interpretation:

The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{HCl}$ is 3.12 which has to be explained

To Explain: The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{HCl}$ is 3.12

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$ and $0.1\text{ mol}$ of $\text{HCl}$ in water.

The pH of this solution is 3.12

pH calculation:

Here, equal moles of strong acid and conjugate base are present, so the solution is equivalent to a solution of $\text{HX}$ and $\text{ACl}$ with a concentration of $0.1\text{ mol}$ for each.

Since $\text{ACl}$ does not affect the pH of the solution, the pH of the solution is due to the weak acid $\text{HX}$.

Conclusion

The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{HCl}$ is 3.12 was explained.

Interpretation Introduction

Interpretation:

The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{NaOH}$ is 13.00 which has to be explained

Concept Introduction:

The relationship between pH and pOH is gives as,

$\text{pH + pOH =14}$

To Explain: The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{NaOH}$ is 13.00

Explanation of Solution

Given data:

A $1.0\text{ L}$ of solution is obtained by dissolving $0.1\text{ mol}$ of the ionic solid $\text{AX}$ and $0.1\text{ mol}$ of $\text{NaOH}$ in water.

The pH of this solution is 13.00

pH calculation:

The given solution has a strong base (Sodium hydroxide) and a weak conjugate base, ${\text{(X}}^{-})$.

The solution is dominated by the strong base.

Since the concentration of $\text{NaOH}$ is $0.1\text{ mol}$, the concentration of hydroxide ion ${\text{[OH}}^{-}]$ is $0.1\text{ mol}$.

Thus,

$\begin{array}{l}\text{pOH}\text{=-log(0}\text{.10)}\\ \text{=1}\text{.00}\end{array}$

The pH is calculated from pOH as follows,

$\begin{array}{l}\text{pH}\text{=14}\text{.00 - pOH}\\ \text{=14}\text{.00 -1}\text{.00}\\ \text{=13}\text{.00}\end{array}$

Conclusion

The pH of the $1.0\text{ L}$ solution prepared by addition of $0.1\text{ mol}$ $\text{AX}$ and $0.1\text{ mol}$ $\text{NaOH}$ is 13.00 was explained

Interpretation Introduction

Interpretation:

The wrong assumption of pH as 1.00 for a solution containing $0.1\text{ mol}$ of $\text{AX}$ and $0.1\text{ mol}$ of $\text{HCl}$ has to be explained.

To explain: The wrong assumption of pH as 1.00 for a solution containing $0.1\text{ mol}$ of $\text{AX}$ and $0.1\text{ mol}$ of $\text{HCl}$

Explanation of Solution

Given data:

Some students mistakenly think that a solution containing $0.1\text{ mol}$ of $\text{AX}$ and $0.1\text{ mol}$ of $\text{HCl}$ should have a pH of 1.00

Explanation for wrong assumption of pH:

This is a solution of equal moles of a strong acid and a weak base.

The students misconception makes them think that the strong acid dominates the behaviour, which is not the real case.

The acid and the base will react with each other and produce an equal number of moles of conjugate acid $\text{HX}$.

Since the conjugate acid is a weak acid, you would not expect it to dissociate completely, so the pH would not be 1.00, which would be the case if it were a strong acid.

Conclusion

The wrong assumption of pH as 1.00 for a solution containing $0.1\text{ mol}$ of $\text{AX}$ and $0.1\text{ mol}$ of $\text{HCl}$ was explained.