Question

Part A

Determine the [OH−] of a 0.23 M solution of NaHCO3.

Part B

Determine the pH of this solution.

Answer 1

Part A:

HCO3- + H2O <----> H2CO3 + OH- (base)

K = [H2CO3][OH-]/[HCO3-]

pOH = ½ x pK -1/2 x log [B]

K_a1of carbonic acid is 4.3 × 10^-7.

Therefore, pKa = −log(4.3 × 10^-7) = 5.3665

pKb = 14 −5.3665 = 8.6335

Therefore, pOH = ½ (8.6335) – ½ (0.23 M)

                                = 4.2

                -log[OH-] = 4.2

                [OH-] = 6.3 x 10^-5 M

Part B :

pH = 14-pOH

     = 14-4.2

    = 9.8