0.653 g of a monoprotic acid (MW= 157 g/mol)is dissolved in water to produce 50.0 mL of a solution with pH = 2.13.Determine the ionization constant of the acid. A)7.9 × 10^-3 B)8.9 × 10^-2 C)6.6 × 10^-4 D)3.9 × 10^-2 E)3.6 × 10^-6

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Quizplus · Accepted Answer

First, calculate the molarity of the acid solution: $ \frac{0.653 \, \text{g}}{157 \, \text{g/mol}} = 0.00416 \, \text{mol} $, and when dissolved in 50.0 mL (or 0.0500 L) of water, the molarity is $ \frac{0.00416 \, \text{mol}}{0.0500 \, \text{L}} = 0.0832 \, \text{M} $. The pH of the solution is 2.13, which means the concentration of $ \text{H}^+ $ ions is $ 10^{-2.13} = 7.41 \times 10^{-3} \, \text{M} $. For a monoprotic acid, the ionization can be represented as $ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- $, and the ionization constant ($ K_a $) is given by $ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} $. Assuming complete ionization of the acid that contributes to the $ \text{H}^+ $ concentration, the initial concentration of the acid before ionization is negligible compared to the change due to ionization, so $ [\text{HA}] \approx 0.0832 \, \text{M} - 7.41 \times 10^{-3} \, \text{M} \approx 0.0832 \, \text{M} $ (since the change is small compared to the initial concentration). Therefore, $ K_a \approx \frac{(7.41 \times 10^{-3})^2}{0.0832} = 6.6 \times 10^{-4} $.

0653 G of a Monoprotic Acid (MW= 157 G/mol)is Dissolved