Question

Question: Consider the titration of 100.0mL of a 0.0500 M solution of the hypothetical weak acid ${\mathbf{H}}_{\mathbf{3}}\mathbf{X}$with 0.100 M KOH. Calculate the pH of the solution under the following conditions

a. before any KOH added

b. after 10.0 mL of 0.100 M KOH has been added

c. after 25.0 mL of 0.100 M KOH has been added

d. after 50.0 mL of 0.100 M KOH has been added

e. after 60.0 mL of 0.100 M KOH has been added

f. after 75.0 mL of 0.100 M KOH has been added

g. after 100.0 mL of 0.100 M KOH has been added

h. after 125.0 mL of 0.100 M KOH has been added

i. after 150.0 mL of 0.100 M KOH has been added

j. after 200.0 mL of 0.100 M KOH has been added

Short answer

a. The pH of the solution before any KOH is added is 2.18 .

b. The pH of the solution after 10.0 mL of 0.100 M KOH has been added is 2.55 .

c. The pH of the solution after 25.0 mL of 0.100 M KOH has been added is 3 .

d. The pH of the solution after 50.0 mL of 0.100 M KOH has been added is 5.

e. The pH of the solution after 60.0 mL of 0.100 M KOH has been added is 6.4 .

f. The pH of the solution after 75.0 mL of 0.100 M KOH has been added is 7 .

g. The pH of the solution after 100.0 mL of 0.100 M KOH has been added is 9.5 .

h. The pH of the solution after 125.0 mL of 0.100 M KOH has been added is 11.63 .

i. The pH of the solution after 150.0 mL of 0.100 M KOH has been added is 12 .

j. The pH of the solution after 200.0 mL of 0.100 M KOH has been added is 12.34 .

Step-by-step solution

Definition of pH

A quantitative measure of the acidity or basicity of aqueous or other liquid solutions is defined as pH.

Calculations

a. The first dissociation of ${\mathrm{H}}_3\mathrm{X}$ is represented as:

${\mathrm{H}}_3\mathrm{X}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{H}}_2{\mathrm{X}}^{-}$

The first acid dissociation constant can be represented as:

role="math" $$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{H}}_2{\mathrm{X}}^{-}\right]}{\left[{\mathrm{H}}_3\mathrm{X}\right]} \\ 1.0\times {10}^{-3}=\frac{{\mathrm{x}}^2}{0.05-\mathrm{x}} \\ {\mathrm{x}}^2+{10}^{-3}\mathrm{x}-5\times {10}^{-5}=0 \\ \mathrm{x}=6.6\times {10}^{-3} \end{gathered}$$

This is the concentration of hydrogen ions in the solution.

The pH can be calculated as:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log (6.6\times {10}^{-3}) \\ \mathbf{pH}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{18} \end{gathered}$$

b. The total volume will be 110 mL. The concentration can be calculated as:

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{n}}{\mathrm{v}} \\ \mathrm{M}=0.0364 \mathrm{M} \end{gathered}$$

The equilibrium expression can be represented as:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}=1\times {10}^{-3} \\ =\frac{\mathrm{x}(0.00909+\mathrm{x})}{(0.0364-\mathrm{x})} \\ {\mathrm{x}}^2+(1.01\times {10}^{-2})\mathrm{x}-3.6\times {10}^{-5}=0 \\ \mathrm{x}=2.8\times {10}^{-3} \end{gathered}$$

The pH of the solution will be:

$$\begin{gathered} \mathrm{pH}=-\log \mathrm{x} \\ \mathrm{pH}=-\log (2.8\times {10}^{-3}) \\ \mathbf{pH}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{55} \end{gathered}$$

c. The molar amount of hydroxide ion initially present can be calculated as:

$$\begin{gathered} \mathrm{n}=\mathrm{M}\times \mathrm{V} \\ =0.1\mathrm{M}\times 25 \mathrm{ml} \\ =2.5 \mathrm{mmol} \end{gathered}$$

The amount of ${\mathrm{H}}_3\mathrm{X}$left can be calculated as:

$5-2.5=2.5 \mathrm{mmol}$

The pH can be calculated as:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}} \\ {\mathrm{pK}}_{\mathrm{a}}=-\log (1.0\times {10}^{-3}) \\ \mathbf{pH}\mathbf{=}\mathbf{3} \end{gathered}$$

d. The number of moles of hydroxide ion can be calculated as:

$$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(50 \mathrm{mL}) \\ \mathrm{n}=5 \mathrm{mmol} \end{gathered}$$

The remaining amount of ${\mathrm{H}}_3\mathrm{X}$will be:

5-5=0

The pH can be calculated as

$$\begin{gathered} \mathrm{pH}=\frac{{\mathrm{pK}}_{\mathrm{a}1}+{\mathrm{pK}}_{\mathrm{a}2}}{2} \\ \mathrm{pH}=\frac{3-\log (1\times {10}^{-7})}{2} \\ \mathbf{pH}\mathbf{=}\mathbf{5} \end{gathered}$$

e. The number of moles of hydroxide ion can be calculated as:

$$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(60 \mathrm{mL}) \\ \mathrm{n}=6 \mathrm{mmol} \end{gathered}$$

The remaining amount of ${\mathrm{H}}_2{\mathrm{X}}^{-}$will be:

$4 \mathrm{mmol} {\mathrm{H}}_2{\mathrm{X}}^{-}$

The molarity is 0.025

The molarity of ${\mathrm{HX}}^{2-}$will be 0.00625

The${\mathrm{K}}_{\mathrm{a}2}$value is$\mathrm{x}=4.0\times {10}^{-7}$

The pH value can be calculated as:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log (4\times {10}^{-7}) \\ \mathbf{pH}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{4} \end{gathered}$$

f. The number of moles of hydroxide ion can be calculated as:

$$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(75 \mathrm{mL}) \\ \mathrm{n}=7.5\mathrm{mmol} \end{gathered}$$

The remaining amount of${\mathrm{H}}_2{\mathrm{X}}^{-}$will be:

$2.5\mathrm{mmol} {\mathrm{H}}_2{\mathrm{X}}^{-}$

The pH value can be calculated as:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log (1.0\times {10}^{-7}) \\ \mathbf{pH}\mathbf{=}\mathbf{7} \end{gathered}$$

g. The initial number of moles of hydroxide ion can be calculated as:

$$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(100 \mathrm{mL}) \\ \mathrm{n}=10\mathrm{mmol} \end{gathered}$$

The remaining amount of ${\mathrm{H}}_2{\mathrm{X}}^{-}$will be:

$0\mathrm{mmol} {\mathrm{H}}_2{\mathrm{X}}^{-}$

The pH value can be calculated as:

$$\begin{gathered} \mathrm{pH}=\frac{{\mathrm{pK}}_{\mathrm{a}2}+{\mathrm{pK}}_{\mathrm{a}1}}{2} \\ \mathrm{pH}=7-\log (1\times {10}^{12}) \\ \mathbf{pH}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{5} \end{gathered}$$

h. The initial number of moles of hydroxide ion can be calculated as:

$$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(125 \mathrm{mL}) \\ \mathrm{n}=12.5 \mathrm{mmol} \end{gathered}$$

The remaining amount of ${\mathrm{H}}_2{\mathrm{X}}^{-}$will be:

$2.5\mathrm{mmol} {\mathrm{H}}_2{\mathrm{X}}^{-}$

The molarity can be calculated as M=0.011M

The pH value can be calculated as:

$$\begin{gathered} \mathrm{pH}=14-\mathrm{pOH} \\ \mathrm{pH}=14+\log \left[{\mathrm{OH}}^{-}\right] \\ \mathbf{pH}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{63} \end{gathered}$$

i. The hydrogen ion concentration can be calculated as:

localid="1649162305642" $$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(150 \mathrm{mL}) \\ \mathrm{n}=15 \mathrm{mmol} \end{gathered}$$

The remaining amount of localid="1649162329658" ${\mathrm{HX}}^{2-}$will be:

localid="1649162393181" $0\mathrm{mmol} {\mathrm{H}}_2{\mathrm{X}}^{-}$

The molarity can be calculated as M=0.02M

The pH value can be calculated as:

localid="1649162422836" $$\begin{gathered} \mathrm{pH}=14-\mathrm{pOH} \\ \mathrm{pH}=14+\log \left[{\mathrm{OH}}^{-}\right] \\ \mathbf{pH}\mathbf{=}\mathbf{12} \end{gathered}$$

j. The hydrogen ion concentration can be calculated as:

localid="1649162446112" $$\begin{gathered} \mathrm{n}=(0.1 \mathrm{M})(200 \mathrm{mL}) \\ \mathrm{n}=20 \mathrm{mmol} \end{gathered}$$

The remaining amount of localid="1649162473284" ${\mathrm{HX}}^{2-}$will be:

5mmol.

The molarity can be calculated as M=0.017M

The pH value can be calculated as:

localid="1649162494516" $$\begin{gathered} \mathrm{pH}=14-\mathrm{pOH} \\ \mathrm{pH}=14+\log \left[{\mathrm{OH}}^{-}\right] \\ \mathbf{pH}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{34} \end{gathered}$$