Chapter 16, Problem 16.112QP

Cacodylic acid is (CH₃)₂AsO₂H. Its ionization constant is 6.4 × 10⁻⁷. (a) Calculate the pH of 50.0 mL of a 0.10 M solution of the acid. (b) Calculate the pH of 25.0 mL of 0.15 M (CH₃)₂AsO₂Na. (c) Mix the solutions in part (a) and part (b). Calculate the pH of the resulting solution.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of final solutions has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation

To find: the $\text{pH}$ of $\text{0}\text{.10}\text{M}\text{Cacodylic acid}$

Answer to Problem 16.112QP

The $\text{pH}$ of final solution is $3.60$

Explanation of Solution

$\begin{array}{l}{\text{CacH}}_{\left(\text{aq}\right)}\to {\text{ Cac}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{+}{\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{0}\text{.10}\text{0 }\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{-x }\text{+x}\text{+x}\\ \text{Final concentration}\left(\text{M}\right)\text{: }\text{0}\text{.10-x}\text{x }\text{x}\\ \\ {\text{K}}_{\text{a}}\text{value for cacodylic acid is 6}\text{.4 ×}{\text{10}}^{\text{-7}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{ ][Cac}}^{\text{-}}\text{ ]}}{\text{[CacH]}}\\ \text{6}\text{.4 ×}{\text{10}}^{\text{-7}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.10-x)}}\approx \frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.10)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [H}}^{\text{+}}\text{] = 2}\text{.5 ×}{\text{10}}^{\text{-4}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(2}\text{.5 ×}{\text{10}}^{\text{-4}}\text{)}\\ \text{pH}\text{=}\text{3}\text{.60}\end{array}$

The concentration of hydrogen ion can be determined using acid dissociation constant.  From the concentration of hydrogen ion, the $\text{pH}$ is calculated by taking negative log of hydrogen ion.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of final solutions has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation

To find the $\text{pH}$ of $\text{0}\text{.15}\text{M}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{AsO}}_{\text{2}}\text{Na}$

Answer to Problem 16.112QP

The $\text{pH}$ of final solution is $9.69$

Explanation of Solution

$\begin{array}{l}{\text{Cac}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ CacH}}_{\left(\text{aq}\right)}\text{+}{\text{OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{0}\text{.15}\text{0 }\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{-x }\text{+x}\text{+x}\\ \text{Equilibrium}\text{concentration}\left(\text{M}\right)\text{: }\text{0}\text{.15-x}\text{x }\text{x}\\ \\ {\text{K}}_{\text{b}}{\text{value for Cac}}^{\text{-}}\text{ is 1}\text{.6 ×}{\text{10}}^{\text{-8}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{6}\text{.4}\text{×}{\text{10}}^{\text{-7}}}\text{=}\text{1}\text{.6 ×}{\text{10}}^{\text{-8}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CacH ][OH}}^{\text{-}}\text{ ]}}{{\text{[Cac}}^{\text{-}}\text{]}}\\ \text{1}\text{.6 ×}{\text{10}}^{\text{-8}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.15-x)}}\approx \frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.15)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [OH}}^{\text{-}}\text{] = 4}\text{.9 ×}{\text{10}}^{\text{-5}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(4}\text{.9 ×}{\text{10}}^{\text{-5}}\text{)}\\ \text{pOH}\text{=}\text{4}\text{.31}\\ \text{pH = 14}\text{.00 - 4}\text{.31 }\\ \text{= 9}\text{.69}\end{array}$

The concentration of hydroxide ion can be determined using base dissociation constant.  From the concentration of hydrogen ion, the $\text{pH}$ is calculated by taking negative log of hydroxide ion and subtracting by fourteen.

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}{\text{NH}}_{\text{3}}\text{ in 10 mL}\text{=}10.\text{0 mL}\text{×}\frac{\text{0}\text{.300 mol}}{\text{1000}\text{mL}}\\ \text{=}3.\text{0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{HCl in 10 mL}\text{=}10.\text{0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

Interpretation Introduction

Interpretation:

The $\text{pH}$ of final solutions has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation

To find the number of moles of $\text{CacH}$ and $\text{CacNa}$

Answer to Problem 16.112QP

The $\text{pH}$ of final solution is $6.07$

Explanation of Solution

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{0}\text{.10M}\text{CacH in 50 mL}\text{=}\text{50}\text{.0 mL}\text{CacH×}\frac{\text{0}\text{.10 mol}\text{CacH}}{\text{1000}\text{mL}}\\ \text{=}\text{5}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{mol CacH}\\ \text{Number}\text{of}\text{moles}\text{of}\text{0}\text{.10M}\text{CacNa in 25 mL}\text{=}\text{25}\text{.0 mLCacNa}\text{×}\frac{\text{0}\text{.150 mol}\text{CacNa}}{\text{1000}\text{mL}}\\ \text{=}\text{3}\text{.8}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{CacNa}\end{array}$

The number of moles of $\text{CacH}$ and $\text{CacNa}$ can be calculated using respective volume and given concentration.

To find the $\text{pH}$ of mixed solution

$\begin{array}{l}\text{We got buffer solution}\text{. The}\text{Henderson-Hasselbalch equation}\text{is}\text{used}\text{to}\text{find}\text{pH}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{{\text{[Cac}}^{\text{-}}\text{]}}{\text{[CacH]}}\\ \\ \text{pH}\text{=}\text{-log(6}\text{.4}\text{×}{\text{10}}^{\text{-7}}\text{)}\text{+}\text{log}\frac{\text{[3}\text{.8}\text{×}{\text{10}}^{\text{-3}}\text{]}}{\text{[5}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{]}}\text{=}\text{6}\text{.07}\end{array}$

The Henderson-Hasselbalch equation can be used to calculate the $\text{pH}$ because the solution acts as buffer system.  From the equation and using acid dissociation constant, the $\text{pH}$ is calculated.