What is the pH of a 0.10 M solution of `Na_(2)SO_(4)? K_(2)(H_(2)SO_(4))=1 xx 10^(-2)`. Assume hydrolysis of `SO_(4)^(2-)" to "HSO_4^-.`

Explanation

As `Na_2SO_4` is a salt of NaOH and `H_2SO_4` it appears that it will not undergo hydrolysis being a salt of a strong acid and a strong base and the pH should, therefore, be 7. But `SO_(4)^(2-)` ions may undergo hydrolysis as `HSO_(4)^(-)` is a weak acid with `K_a= 1 xx 10^(-2).` The hydrolysis equilibrium is `SO_(4)^(2-)+H_(2)O Leftrightarrow HSO_(4)^(-)+OH^-`
As `Na_(2)SO_(4)` is a strong electrolyte, `[SO_4^(2-)]= [Na_2SO_4]= 0.10.`
`pH =1/2 [pK_w +pK_a +log a]`
`=1/2 {-log 10^(-14) +(-log 10^(-2))+log (0.10)}`
=7.5