Question

The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.04. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.

Answer 1

Let,

NaC6H4ClO = NaC6H4ClO^-

Now,

C6H4ClO^- + H2O    HC6H4ClO + OH^-

Kb = [HC6H4ClO] [OH^-] / [C6H4ClO^-]

Ka for HC6H4ClO = 6.6 x 10^-10

Kb = kw / Ka
= (1.0 x 10^-14) / ( 6.6 x 10^-10)
= 1.5 x 10^-5

pH = 11.04

So, pOH = 14 - pH

pOH = 14 - 11.04

- log  [OH^-] = 2.96

log  [OH^-] = - 2.96

[OH^-] = 10^{- 2.96}

[OH^-] = 1.1 x 10^-3 M

So, [HC6H4ClO] =  [OH^-] = 1.1 x 10^-3 M

Now,

Kb = [HC6H4ClO] [OH^-] / [C6H4ClO^-]

1.5 x 10^-5 = (1.1 x 10^-3) (1.1 x 10^-3) / [C6H4ClO^-]

[C6H4ClO^-]  = (1.1 x 10^-3) (1.1 x 10^-3) / (1.5 x 10^-5)

[C6H4ClO^-]  = 0.08 M