Question

a) Calculate the pH of a 0.045 M potassium chlorite solution. (The pK_a for chlorous acid is 1.92)

b) Calculate the pH of a solution that is 0.35 M hydroxylamine and 0.50 M hydroxylammonium chloride, (The pK_b for hydroxylamine is 7.96)

Answer 1

Ans. #A. Potassium chlorite (salt) dissociates into chlorite ion (conjugate base of chlorate acid) and K⁺ (spectator ion). The chlorite ion, acting as conjugate base, accepts a proton water to form chlorous acid.

            KClO₂(s) -------------> K⁺ + ClO₂^-              - reaction 1

            ClO₂^- + H2O --------> HClO₂ + OH^-         - reaction 1

Given, pKa of chlorous acid = 1.92

Now, pKb (base dissociation constant for conjugate base) = 14.00 – pKa

= 14.00 – 1.95

= 12.05

Using the formula, pKb = -log Kb

            Or, Kb = antilog (-pKb) = antilog (- 12.05) = 8.913 x 10^-13

# Create an ICE table for reaction 2 as shown in figure-

Now,

            Kb = (HClO₂) (OH^-) / (ClO₂^-)                        - all concentrations at equilibrium

Or, 8.913 x 10^-13 = (X) (X) / (0.045 -X)

Or, (8.913 x 10^-13) x (0.045 -X) = X²

Or, X² + (8.913 x 10^-13)X – (4.011 x 10^-14) = 0

Solving the quadratic equation, we get following two roots-

            X1 = 2.003 x 10^-7      ; X2 = - 2.003 x 10^-7

Since concentration can’t be negative, reject X2.

            Hence, X = 2.003 x 10^-7

Therefore, at equilibrium [OH^-] = X = 2.003 x 10^-7 M

            pOH = -log [OH^-] = - log (2.003 x 10^-7) = 6.98

Now, using the formula, pH + pOH = 14.00

            pH = 14.00 - pOH = 14.00 – 6.98 = 7.02

Therefore, pH of the solution = 7.02     

#B. A solution of hydroxylamine (NH₂OH) and hydroxylammonium chloride [ (NH₃OH)Cl ; HCl salt of the base hydroxylamine] acts as a buffer system.

Hydroxylamine act as base. hydroxylammonium chloride acts as conjugate acid.

Using Henderson- Hasselbalch equation for a (base- conjugate acid) buffer system–

            pOH = pKb + log ([conjugate acid] / [Base])              - equation 1

Putting the values in above equation-

            pOH = 7.96 + log (0.50 M / 0.35 M)

            or, pOH = 7.96 + log 1.4287 = 7.96 + 0.155 = 8.12

Hence, pOH = 8.12

Now, using the formula, pH + pOH = 14.00

            pH = 14.00 - pOH = 14.00 – 8.12 = 5.88

Therefore, pH of the solution = 5.88