Final answer:
To calculate the pH for a 0.142M solution of NaHA, we need to consider the second ionization constant (Ka2) for the diprotic acid H2A. By setting up and solving a quadratic equation using the Ka2 value, we can find the equilibrium concentration of H+ ions and determine the pH of the solution.
Step-by-step explanation:
To calculate the pH for a 0.142M solution of NaHA, we need to consider the ionization of the diprotic acid, H2A. The ionization constants, Ka1 = 2.3×10^-4 and Ka2 = 3.8×10^-12, represent the equilibrium constants for the dissociation of H2A. In this case, we can assume that the dissociation is negligible for the first ionization (Ka1) because the concentration of H2A is much greater than its equilibrium concentration. Therefore, we only need to consider the second ionization (Ka2).
The second ionization of H2A can be represented by the equation: H2A(aq) ⇌ HA^-(aq) + H^+(aq).
We can use the equation for Ka to calculate the equilibrium concentration of HA^-(aq) and H^+(aq). Since Ka2 = [HA^-][H^+]/[H2A], we can rearrange the equation to solve for [H^+]. Given that the concentration of NaHA is 0.142M, we can assume that [H2A] = 0.142M.
Substituting the values into the equation, we have:
3.8×10^-12 = x^2 / (0.142 - x)
where x represents the concentration of H^+ ions. By solving this quadratic equation, we can find the equilibrium concentration of H^+ and use it to calculate the pH of the solution.