Question

Consider the titration of $\mathbf{100}\mathbf{.}\mathbf{0}\mathit{m}\mathit{L}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{200}\mathit{M}\mathbf{}\mathbf{ }\mathit{a}\mathit{c}\mathit{e}\mathit{t}\mathit{i}\mathit{c}\mathbf{}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathbf{}\mathbf{ }\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathbf{}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathit{M}\mathbf{}\mathbf{ }\mathit{K}\mathit{O}\mathit{H}\mathbf{.}$Calculate the pH of the resulting solution after the following volumes of HCl have been added.

$$\begin{gathered} \mathit{a}\mathbf{.}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{L} \\ \mathit{b}\mathbf{.}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{L} \\ \mathit{c}\mathbf{.}\mathbf{}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{L} \\ \mathit{d}\mathbf{.}\mathbf{}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{L} \\ \mathit{e}\mathbf{.}\mathbf{}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{L} \end{gathered}$$

Short answer

1. The pH of the resulting solution is 2.72
2. The pH of the resulting solution is 4.26
3. The pH of the resulting solution is 4.74
4. The pH of the resulting solution is 5.22
5. The pH of the resulting solution is 8.79
6. The pH of the resulting solution is 12.14

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

Calculation of pH

a.

$\begin{array}{c}x=\left[H^{+}\right]\\ x=1.9\times {10}^{-3}M\\ pH=-log\left[H^{+}\right]\\ pH=-log\left(1.9\times {10}^{-3}\right)\\ pH=2.72\end{array}$

b. The added reacts completely with acetic acid.

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}$

Number of moles of acetic acid present =0.02 mol

Number of moles of${\mathrm{OH}}^{-}$ added =0.005 mol

Number of moles of acetic acid left = 0.015 mol

Number of moles of acetate ions formed = 0.005 mol

If the final volume is V,

$\begin{array}{c}pH=p{K}_a+log\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}\\ pH=-log\left(1.8\times {10}^{-5}\right)+log\frac{\left(0.005/V\right)}{\left(0.015/V\right)}\\ pH=4.74+(-0.477)\\ pH=4.26\end{array}$

c.

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}$

Number of moles of acetic acid present =0.02 mol

Number of moles of${\mathrm{OH}}^{-}$ added =0.01 mol

Number of moles of acetic acid left = 0.01 mol

Number of moles of acetate ions formed = 0.01 mol

If the final volume is V,

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}\\ \mathrm{pH}=-\log \left(1.8\times {10}^{-5}\right)+\log \frac{\left(0.01/\mathrm{V}\right)}{\left(0.01/\mathrm{V}\right)}\\ \mathrm{pH}=4.74\end{array}$

d.

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}$

Number of moles of acetic acid present =0.02 mol

Number of moles of${\mathrm{OH}}^{-}$ added =0.015 mol

Number of moles of acetic acid left = 0.005 mol

Number of moles of acetate ions formed = 0.015 mol

If the final volume is V,

$\begin{array}{l}pH=p{K}_a+log\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}\\ pH=-log\left(1.8\times {10}^{-5}\right)+log\frac{\left(0.015/V\right)}{\left(0.005/V\right)}\\ pH=4.74+0.48\\ pH=5.22\end{array}$

e.

Number of moles of acetic acid present =0.02 mol

Number of moles of${\mathrm{OH}}^{-}$ added =0.02 mol

Number of moles of acetic acid left = 0.0 mol

Number of moles of acetate ions formed = 0.02 mol

$\begin{array}{c}\mathrm{x}=6.1\times 10{}^{-6}\mathrm{M}\\ \mathrm{pOH}=-\log (6.1\times 10{}^{-6})\\ \mathrm{pOH}=5.21\\ \mathrm{pH}=14-5.21\\ \mathrm{pH}=8.79\end{array}$

f.

Number of moles of${\mathrm{OH}}^{-}$ added =0.025 mol

Number of moles of acetate ions formed = 0.02 mol

Number of excess ${\mathrm{OH}}^{-}$moles from KOH= 0.005 mol

$\begin{array}{c}\mathrm{pOH}=-\log (0.014)\\ \mathrm{pOH}=1.86\\ \mathrm{pH}=14-1.86\\ \mathrm{pH}=12.14\end{array}$