7.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

NaNO2is the salt of a strong base (NaOH) and a weak acid (HNO2).

NO2-+H2OHNO2+OH-Kh=[HNO2][OH-][NO2-]KwKα=10-144.5×10-4=.22×10-10

Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species present in the solution will be:

[NO2-]=.04-x;0.04[HNO2]=x[OH-]=xKh=x20.04=0.22×10-10x2=.0088×1010x=0.93×10-5[OH-]=0.093×10-5M[H2O+]=10-14.093×10-5=10.75×10-9 MpH=-log(10.75×10-9)=7.96Therefore, degree of hydrolysis =x0.04=.093×10-5.04=2.325×10-5