Question

What is the pH of a 0.126 M monoprotic acid whose Ka is 7.145 × 10−3?

Answer 1

Ans. Create an ICE table for the monoprotic acid (AH) as shown in picture.

Now,

Acid dissociation constant, Ka = [A^-] [H3O⁺] / [AH]        , where all concentrations are those at equilibrium.

Putting the value in above equation-

            7.145 x 10^-3 = (X) (X) / (0.126 - X)

            Or, 0.007145 x (0.126 - X) = X²

            Or, X² + 0.007145X – 0.00090027 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.0266              , X2 = -0.03379

Since concertation can’t be negative, reject X2.

Hence, X = 0.0266

Therefore, [H₃O⁺] at equilibrium = X = 0.0266 M

Now, using the formula      pH = -log [H₃O⁺]

            pH = -log (0.0266) = 1.58

Therefore, pH of the solution = 1.58