Question

Consider the titration of 100.0mL of asolution of an acid HA(${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$) with . Calculate the pH for the following conditions.
a. before any NaOH has been added
b. after 5.00 mL NaOH has been added
c. at the stoichiometric point

Short answer

The pH for the following conditions:

a)Before anyNaOHhas been added is 6.6.

b) After 5.00 mL has been added is 9.30.

c) At the stoichiometric point is 4.47.

Step-by-step solution

Subpart (a) Determination of pH for the condition before any NaOH has been added

As theHA solution is very dilute, water autoionizationshould also be addressed.

$$\begin{gathered} \mathrm{HA}\to {\text{H}}^{+}+{\mathrm{A}}^{-} \\ 2{\mathrm{H}}_2\mathrm{O}\to {\text{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-} \\ \left[{\mathrm{H}}^{+}\right]=\sqrt{\left({\mathrm{K}}_{\mathrm{a}}{\left[\mathrm{HA}\right]}_0+{\mathrm{K}}_{\mathrm{w}}\right)} \\ =\sqrt{\left(5.0\times {10}^{-10}\times 1.00\times {10}^{-4}\right)+1.0\times {10}^{-14}} \\ =2.45\times {10}^{-7}\text{M} \end{gathered}$$

The following equation can be used to determine the pH of a solution.

role="math" localid="1663761139698" $$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left[2.45\times {10}^{-7}\text{M}\right] \\ =6.6 \end{gathered}$$

Therefore, the pH is 6.6.

Subpart (b) Determination of pH for the condition after 5.00 mL of NaOH has been added.

$$\begin{gathered} \text{Left Concentration o}\mathrm{f}\mathrm{HA}=\frac{\left(1.00\times {10}^{-5}-5.0\times {10}^{-6}\right)\text{mol}}{105.0\text{mL}}\times 1000\text{mL} \\ =4.76\times {10}^{-5}\text{M} \end{gathered}$$Here, the number of moles of acid HA:,

$$\begin{gathered} \mathrm{HA}=\frac{1.00\times {10}^{-4}\text{mol}}{1000\text{mL}}\times 100.0\text{mL} \\ =1.00\times {10}^{-5}\text{mol} \end{gathered}$$

The number of moles added toNaOH:

$$\begin{gathered} \mathrm{NaOH}=\frac{1.00\times {10}^{-3}\text{mol}}{1000\text{mL}}\times 5.0\text{mL} \\ =5.00\times {10}^{-6}\text{mol} \end{gathered}$$

Also,

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

$$\begin{gathered} \mathrm{pH}=-\log \left[5.0\times {10}^{-10}\right] \\ \text{= - ( - 10) + ( - 0.7)} \\ \text{= 9.3} \end{gathered}$$

Hence, pH =9.3

Subpart (c) Determination of pH for the condition at the stoichiometric point

Let’s convert into:

Initial		-	-
eqm		x	x

As,

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{HA}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{A}}^{-}\right]}$

Putting the values in the equation,

$2.0\times {10}^{-5}=\frac{{\mathrm{x}}^2}{\left(9.09\times {10}^{-5}-\mathrm{x}\right)}$

so,

$$\begin{gathered} \mathrm{pH}=-\log \left[\mathrm{x}\right] \\ \mathrm{pH}=-\log \left[3.4\times {10}^{-5}\text{M}\right] \\ \mathrm{pH}=-(-5)+(-0.53) \\ \mathrm{pH}=4.47 \end{gathered}$$

Therefore, pH is 4.47.