Question

The following questions A24 - A26 relate to 100 ml of 0.0150 M solution of benzoic acid
(C6H3COOH). Ka(C6H3COOH) = 6.4 x 10^-5.

What is the pH of the solution after the addition of 1 x 10^-3 moles of NaOH? You may assume no volume change to the solution upon addition of the NaOH.

Answer 1

Answer:

PH of the weak acid: approximately 2.513.

PH of the buffer solution: approximately 4.495.

Explanation:

The Ka value of benzoic acid is much smaller than 1. Benzoic acid will dissociate but only partially when dissolved in water. Construct a RICE table for this process. Let the equilibrium of $\rm H^{+}$ be $x\; \rm mol\cdot L^{-1}$. Note that $x \ge 0$.

$\displaystyle \begin{array}{c|ccccc}\textbf{R}&\mathrm{C_6H_5COOH} & \rightleftharpoons & \mathrm{C_6H_5COO^{-}} & + &\mathrm{H^{+}}\\\textbf{I} & 0.015 & & 0 & & 0\\\textbf{C} & -x & & +x & & +x\\\textbf{E} & 0.015-x & & x & & x\end{array}$

$\displaystyle \frac{[\mathrm{C_6H_5COO^{-}}]\cdot [\mathrm{H^{+}}]}{[\mathrm{C_6H_5COOH}]} = \mathrm{pK}_{a}$.

$\displaystyle \frac{x^{2}}{0.015 - x} = 6.4\times 10^{-5}$.

Solve this quadratic equation for $x$:

$x^{2}+ 6.4\times 10^{-5}\;x - 6.4\times 10^{-5}\times 0.150 = 0$.

$\displaystyle x = \frac{-6.4\times 10^{-5} \pm \sqrt{(6.4\times 10^{-5})^{2} - 4\times (- 6.4\times 10^{-5}\times 0.150)}}{2}$.

Take only the non-negative root. $x \approx 0.00306655$.

$\rm [H^{+}] = 0.00306655\; mol\cdot L^{-1}$.

$\displaystyle \mathrm{pH} = -\log_{10}{[\mathrm{H^{+}}]} = 2.513$.

Each benzoic acid contains only one carboxyl group $\mathrm{-COOH}$. Benzoic acid is thus a monoprotic acid. Each mole of the acid will react with only one mole of $\rm NaOH$. The 100 mL solution initially contains $1.50\times 10^{-3}$ moles of benzoic acid. The $1\times 10^{-3}$ moles of $\rm NaOH$ will neutralize only part of the acid. The solution will eventually contain $1\times 10^{-3}$ moles of $\mathrm{C_6H_5COO^{-}}$ (from the salt $\mathrm{C_6H_5COONa}$) and $0.50\times 10^{-3}$ moles of $\mathrm{C_6H_3COOH}$.

Both the acid $\mathrm{C_6H_5COOH}$ and the conjugate base of the acid $\mathrm{C_6H_5COO^{-}}$ exist in large amounts in the solution. Apply the Henderson-Hasselbalch equation for weak acid buffers to find the pH of this buffer solution.

$\mathrm{pK}_{a} = -\log_{10}{\mathrm{K}_{a}} \approx 4.19382$ for benzoic acid.

$\begin{aligned}\mathrm{pH} &= \mathrm{pK}_{a} + \log{\frac{{[\text{Conjugate Base}]}}{[\text{Weak Acid}]}} \\ &= \mathrm{pK}_{a} + \log{\frac{{[\mathrm{C_6H_5COO^{-}}]}}{[\mathrm{C_6H_5COOH}]}}\\ &= 4.19382 + \log{\frac{0.01}{0.005}}\\ &\approx 4.495 \end{aligned}$.