Question

Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. ($K_a$ of acetic acid = $1.8\times {10}^{-5}$)

• A. 5.6
• B. 4.74
• C. 8.62
• D. 12.8

Answer 1

The correct option is B

4.74

$C{H}_{3}COOH+H_{2}O\to H_3{O}^{+}C{H}_{3}CO{O}^{-}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}=1.8\times {10}^{-5}\text{Before treatment}\left[H_3{O}^{+}\right]=\left[C{H}_{3}CO{O}^{-}\right]=x\left[C{H}_{3}COOH\right]=0.10K_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}=1.8\times {10}^{-5}1.8\times {10}^{-5}=\frac{x\times x}{0.10}\frac{x^2}{0.10}=1.8\times {10}^{-5}{x}^2=1.8\times {10}^{-6}x=\left[H_3{O}^{+}\right]=1.3\times {10}^{-3}pH=2.87$

After treatment : Half the acid is neutralized and half remains both solute are now in 75.0ml of solution.

$=\left[C{H}_{3}CO{O}^{-}\right]=\frac{0.050\times 50}{75}y=0.033y.\left[C{H}_{3}COOH\right]=\frac{0.050\times 50}{75}y=0.033K_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}=1.8\times {10}^{-5}=1.8\times {10}^{-5}=\left[H_3{O}^{+}\right]\times \frac{0.033y}{0.033y}$

$\left[H_3{O}^{+}\right]=1.8\times {10}^{-5}$ M

thus pH = 4.74