SAT Chemistry

Answer: D, 11.48.

The solution is as follows:

The equilibrium equation is K_b=[NH₄][OH^-]/[NH₃]

Since NH₃ will not fully dissociate, the concentrations of OH^- and HN₃ are unknown (x) and the concentration of NH₃ at equilibrium will be 0.5 - x.

K_b = [x][x]/[0.500-x] Assuming x is so small it’s negligible, that yields;
K_b = x²/[0.500]
(1.78 * 10^-5) = x²/[0.500]
(1.78 * 10^-5) * 0.500 = x²
x= .003M which is the concentration of OH^-

Using the equilibrium equation for water, we have H₂O ® H⁺ + OH^-

The equilbrium value, K_w is 10^-14

K_w = [H⁺][OH^-]
* 10^-14 = [H⁺][0.003]
[H⁺] = 3.33*10^-12

pH = -log [H⁺] = -log (3.33 * 10^-12) = 11.48