Question

If the Ka of a monoprotic weak acid is 3.0 × 10-6, what is the pH of a 0.45 M solution of this acid?

Answer 1

Concentration of acid is 0.45 m ka of acid is 3.0 times 106-6 acid is 3.0 times 10^-6 Acid hydrolysis reaction is, Now, substitute these concentrations and the value of ka into the equilibrium - constant equation: ka = [A^-][H_3O^-]/[HA] 3.0 times 10^-6 = x times x/0.45 - x Assume x is much smaller than 0.45, so (0.45 - x) = 0.45 x^2 = 3.0 times 10^-6 times 0.45 x = (1.35 times 106-6)^1/2 x = 1.16 times 10^-3 [H_3O^+] = 1.16 times 10^-3 M p^H = -log[H_3O^+] = -log(1.16 times 11106-3) = 2.93 hence, the p^H monoprotic weak acid is 2.93