Calculate the pH of a 0.020 M carbonic acid solution,H₂CO₃(aq),that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7and K_a2 = 5.6 × 10^-11. A)1.70 B)4.03 C)6.37 D)10.25

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Given the stepwise dissociation constants for carbonic acid (H2CO3), $K_{a1} = 4.3 \times 10^{-7}$ and $K_{a2} = 5.6 \times 10^{-11}$, we can assume that the first dissociation step contributes significantly more to the acidity of the solution than the second step due to the much higher $K_{a1}$ value. Therefore, we can approximate the pH of the solution by considering only the first dissociation step.The dissociation of carbonic acid can be represented as:$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$Using the acid dissociation constant, $K_{a1}$, we can set up the expression:$$K_{a1} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]}$$Assuming that the concentration of $H^+$ and $HCO_3^-$ produced is $x$ and the initial concentration of $H_2CO_3$ is 0.020 M, the equation simplifies to:$$4.3 \times 10^{-7} = \frac{x^2}{0.020 - x}$$Given that $x$ is very small compared to 0.020 M, we can approximate the denominator to 0.020 M, leading to:$$x^2 = 4.3 \times 10^{-7} \times 0.020$$$$x^2 = 8.6 \times 10^{-9}$$$$x = \sqrt{8.6 \times 10^{-9}}$$$$x \approx 9.27 \times 10^{-5}$$The concentration of $H^+$ ions, $x$, is approximately $9.27 \times 10^{-5}$ M. To find the pH:$$pH = -\log[H^+]$$$$pH = -\log(9.27 \times 10^{-5})$$$$pH \approx 4.03$$Therefore, the pH of the 0.020 M carbonic acid solution is approximately 4.03.

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