7.64 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

It is given that Ka for ClCH2COOH is 1.35 × 10-3. Kα=2α=Kαc=1.35×10-30.1α=1.35×10-2=0.116[H+]==0.1×0.116=.0116pH=-log[H+]=1.94ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH. CICH2COO-+H2OCICH2COOH+OH-Kb=[CICH2COOH][OH-][CICH2COO-]Kb=KwKαKb=10-141.35×10-3=0.740×10-11Also, Kb=x20.10.740×10-11=x20.1    (where x is the concentration of OH- and CICH2COOH)0.740×10-11=x20.10.074×10-11=x2x2=0.74×10-12x=0.86×10-6[OH-]=0.86×10-6[H+]=Kwo.86×10-6=10-140.86×10-6[H+]=1.162×10-8pH=-log [H+]=7.94