The ionization constant of nitrous acid is `4.5xx10^(-4)`. Calculate the `pH` of `0.04 M` sodium nitrite solution and also its degree of hydrolysis.

Explanation

`pH=1/2pK_w+1/2pK_a _ log c`
`pK_a=-log K_a -log (4.5xx10^(-4))`
=-(0.65-4)=3.35
`therefore pH =1/2(14)+1/2(3.35)+log 0.04`
=7+1.675 -1.39 =7.285
`h=sqrt(K_w/K_(a .c))`
`=sqrt((1xx10^(-14))/(4.5xx10^(-4)xx0.04))=2.35xx10^(-5)`