Question

A 0.316 M solution of NaY has a pH of 10.21. If HY is a weak acid, what is the pKa for this weak acid?

Answer 1

1.20×10^-7

Explanation

Y- from NaY is hydrolysed by water and the equillibrium is

Y- + H2O ------> HY + OH-

Kb = [ OH- ] [ HY ]/[ Y- ]

at equillibrium

[ OH- ] = x

[ HY ] = x

[ Y- ] = 0.316- x

Therefore,

Kb = x^2/(0.316 - x )

Now ,calculate [ OH- ] from pH

pH = 10.21

pOH = 14 -10.21 = 3.79

pOH = - log [ OH- ]

-log[OH] = 3.79

[ OH- ] = 1.62×10^-4M

[ OH-] = x

so , x = 1.62 × 10^-4

substitute x value in Kb expression

Kb = (1.62 ×10^-4)^2/(0.316 - 1.62 × 10^-4)

= 2.62 × 10^-8/0.3158

= 8.30 × 10^-8

Ka = Kw/Kb

where, Kw is ionic product of water , 1.00×10^-14

Therefore,

Ka = 1.00×10^-14/8.30×10^-8

= 1.20×10^-7