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)
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?
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