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

PH of the weak acid: approximately 2.513.

PH of the buffer solution: approximately 4.495.

Step-by-step 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}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 (x^(2))/(0.015 - x) = 6.4* 10^(-5)`.

Solve this quadratic equation for
`x`:

`x^(2)+ 6.4* 10^(-5)\;x - 6.4* 10^(-5)* 0.150 = 0`.

`\displaystyle x = \frac{-6.4* 10^(-5) \pm \sqrt{(6.4* 10^(-5))^(2) - 4* (- 6.4* 10^(-5)* 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* 10^(-3)` moles of benzoic acid. The
`1* 10^(-3)` moles of
`\rm NaOH` will neutralize only part of the acid. The solution will eventually contain
`1* 10^(-3)` moles of
`\mathrm{C_6H_5COO^(-)}` (from the salt
`\mathrm{C_6H_5COONa}`) and
`0.50* 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{(0.01)/(0.005)}\\ &\approx 4.495 \end{aligned}`.