Question

What is [H3O+] and ph in a 0.10 M solution of HCN at 25 degrees celsius (ka for HCN= 4.0x10^-10) show steps

Answer 1

HCN(aq) <----------> CN^-(aq)   +   H⁺(aq)   ; K_a = {[H⁺]*[CN^-]}/[HCN] = 4*10^-10

Initially, [HCN] = 0.1 M ; [H⁺] = [CN^-] = 0 M

Let at eqb., [HCN] = (0.1-x) M ; [H⁺] = [CN^-] = x M

Thus, 4*10^-10 = x²/(0.1-x)

or, x² + (4*10^-10)x - 4*10^-11 = 0

or, x = 6.324*10^-6 M = [H⁺] = [H₃O⁺]

Now, pH = -log[H⁺] = -logx = 5.199