Calculate the pH of a solution produced by mixing 4.3 cm³ of 0.100 mol dm^-3 potassium hydroxide, KOH, solution to 50.0 cm³ of 0.136 mol dm^-3 propanoic acid, C₂H₅COOH, solution. The pK_a of propanoic acid is 4.87.

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Quizplus · Accepted Answer

To find the pH of the solution, first, we need to calculate the moles of KOH and propanoic acid. For KOH: $4.3 \, \text{cm}^3 \times 0.100 \, \text{mol dm}^{-3} = 0.00043 \, \text{mol}$. For propanoic acid: $50.0 \, \text{cm}^3 \times 0.136 \, \text{mol dm}^{-3} = 0.0068 \, \text{mol}$. When KOH is mixed with propanoic acid, it neutralizes some of the acid, forming water and the propanoate ion. The reaction consumes all of the KOH, leaving $0.0068 - 0.00043 = 0.00637 \, \text{mol}$ of propanoic acid unreacted. The concentration of the propanoic acid in the final solution (which has a volume of $4.3 + 50.0 = 54.3 \, \text{cm}^3$ or $0.0543 \, \text{dm}^3$) is $0.00637 \, \text{mol} / 0.0543 \, \text{dm}^3 = 0.117 \, \text{mol dm}^{-3}$. Using the Henderson-Hasselbalch equation, $ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) $, where $[\text{A}^-]$ is the concentration of the propanoate ion (which is equal to the initial moles of KOH, $0.00043 \, \text{mol}$, divided by the total volume in dm$^3$, $0.0543$, giving $0.00792 \, \text{mol dm}^{-3}$) and $[\text{HA}]$ is the concentration of the unreacted propanoic acid. Plugging in the numbers: $ \text{pH} = 4.87 + \log \left( \frac{0.00792}{0.117} \right) $, which approximates to a pH of 3.67.

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