Questions: Find the PH of a 0.080 M HH2CO3 solution. For H2CO3, Ka1=4.3 × 10^-7 and Ka2=5.6 × 10^-11

Solution Steps

We need to find the pH of a 0.080 M solution of carbonic acid (\(\mathrm{H}_2\mathrm{CO}_3\)). Carbonic acid is a diprotic acid, meaning it can donate two protons. We are given the first dissociation constant \(K_{a1} = 4.3 \times 10^{-7}\) and the second dissociation constant \(K_{a2} = 5.6 \times 10^{-11}\).

Since \(K_{a1} \gg K_{a2}\), the first dissociation will contribute most to the hydrogen ion concentration. The dissociation of \(\mathrm{H}_2\mathrm{CO}_3\) can be represented as: \[ \mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \] The equilibrium expression is: \[ K_{a1} = \frac{[\mathrm{H}^+][\mathrm{HCO}_3^-]}{[\mathrm{H}_2\mathrm{CO}_3]} \] Assuming \(x\) is the concentration of \(\mathrm{H}^+\) produced, we have: \[ K_{a1} = \frac{x^2}{0.080 - x} \approx \frac{x^2}{0.080} \] Solving for \(x\): \[ x^2 = K_{a1} \times 0.080 = 4.3 \times 10^{-7} \times 0.080 \] \[ x^2 = 3.44 \times 10^{-8} \] \[ x = \sqrt{3.44 \times 10^{-8}} = 5.865 \times 10^{-5} \]

The pH is calculated using the concentration of \(\mathrm{H}^+\): \[ \text{pH} = -\log_{10}(x) = -\log_{10}(5.865 \times 10^{-5}) \] \[ \text{pH} \approx 4.232 \]

Final Answer