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.

The ionization constant of nitrous acid is 4.5xx10^(-4). Calculate the

1.	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.
Answer» Solution :Sodium nitrite is a SALT of weak acid , strongbase. 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^(-)),("Initialconc.",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`