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

Sodium nitrite is a salt of weak acid , strong base. Hence,
`K_(h)=2.22xx10^(-11)=K_(w)//K_(a)=10^(-14)//(4.5xx10^(-4))`
`h=sqrt(K_(h)//c)=sqrt(2.22xx10^(-11)//0.04)=sqrt(5.5xx10^(-10))=2.36xx10^(-5)`
`{:(,NO_(2)^(-),+,H_(2)O,hArr,HNO_(2) ,+,OH^(-)),("Initial conc.",c,,,,,,),("After hydrolysis",c-ch,,,,ch,,ch):}`
`[OH^(-)]=ch=0.04xx2.36xx10^(-5)=9.44xx10^(-7)`
`pOH=-log (9.44xx10^(-7))=7-0.9750=6.03`
`pH = 14 - pOH = 14 - 6.03 = 7.97`.
or directly, as `NaNO_(2)` is a salt of weak acid and strong base,
`pH=(1)/(2) [pK_(w)+pK_(a)+logC]`
`pK_(w)=-log K_(w)=14`
`pK_(a)=-logK_(a)=-log (4.5xx10^(-4))=-(0.65-4)=3.35`
`log C = log 0.04 = log 4 xx 10^(-2) = - 2 + 0.6021 = - 1.3979 ~=-1.40`
`:. pH = (1)/(2) [14+3.35-1.40]=7.97`