In a solution prepared by dissolving 0.200 mole of a weak acid HX in enough water to make 1.00 L of solution, the pH is observed to 5.350. What is the Ka for this acid?
To determine the Ka for the weak acid HX, we need to use the equation for the dissociation of the acid in water:
HX(aq) ⇌ H+(aq) + X-(aq)
The equilibrium constant expression is:
Ka = [H+(aq)][X-(aq)]/[HX(aq)]
Given that the pH of the solution is 5.350, we can calculate the concentration of H+ ions using the equation:
pH = -log[H+]
[H+] = 10^(-pH) = 10^(-5.350) = 2.97 x 10^(-6) M
Since HX is a weak acid, we can assume that the concentration of HX is approximately equal to the initial concentration of the acid:
[HX(aq)] = 0.200 M
Now, let's assume that the concentration of X- formed from the dissociation of HX is negligible compared to [HX]. This enables us to simplify the equation:
Ka ≈ [H+(aq)][X-(aq)]/[HX(aq)]
Ka ≈ [H+][X-]/[HX]
Ka ≈ [H+]^2/[HX]
Plugging in the values we know:
Ka ≈ (2.97 x 10^(-6))^2 / 0.200
Ka ≈ 8.83 x 10^(-14)