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?

Answers (1)
M manish

We have,
Ionisation constant of chloroacetic acid(K_a) is 1.35\times 10^{-3}
The concentration of acid = 0.1 M
Ionisation if acid, = 
ClCH_2COOH\rightleftharpoons ClCH_2COO^-+H^+

We know that,
\Rightarrow K_a =\frac{[ClCH_2COO^-][H^+]}{[ClCH_2COOH]}....................(i)
As it completely ionised
[ClCH_2COO^-]=[H^+]

Putting the values in eq (i)
1.35\times 10^{-3} = \frac{[H^+]^2}{0.02}
[H^+] = \sqrt{1.35\times 10^{-3}\times 0.02}=1.16\times 10^{-2}
Therefore, pH of the solution = -\log (1.16\times 10^{-2})
                                                  = 2-\log(1.16)
                                                  = 1.94

Now,  

0.1 M  ClCH_2COONa (sod. chloroacetate) is basic due to hydrolysis-

ClCH_2COO^-+ \: H_2O\rightleftharpoons CH_2ClCOOH+OH^-

For a salt of strong base+strong acid

\\pH=7+\frac{pK_a+logC}{2}=7+\frac{2.87+log0.1}{2}\\pH=7.94

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