Question

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?

Answer 1

It is given that K_a for ClCH₂COOH is 1.35 × 10^–3.

`=> "K"_"a" = "c" alpha^2`

`therefore alpha =  sqrt("K"_"a"/"c")`

`= sqrt((1.35 xx 10^(-3))/0.1)`         (∴ concetration of acid = 0.1 m)

`alpha = sqrt(1.35 xx 10^(-2))`

= 0.116

`therefore ["H"^+] = "c"  alpha = 0.1 xx 0.116`

= .0116

`therefore ["H"^+] = "c"  alpha = 0.1 xx 0.116`

= .0116

`=>"pH" = - log["H"^+] = 1.94`

ClCH₂COONa is the salt of a weak acid i.e., ClCH₂COOH and a strong base i.e., NaOH.

\[\ce{ClCH_2COO^- + H_2O <-> ClCH_2COOH + OH^-}\]

`"K"_"b" = (["ClCH"_2"COOH"]["OH"^(-)])/["ClCH"_2"COO"^(-)]`

`"K"_"h" = "K"_"w"/"K"_"a"`

`"K"_"h" = 10^(14)/(1.35 xx 10^(-3))`

`= 0.740 xx 10^(-11)`

Also `"K"_"h" = x^2/0.1`  (where x is the concentration of `"OH"^(-) and "ClCH"_2"COOH")`

` 0.740 xx 10^(-11) = x^2`

`0.074 xx 10^(-11) = x^2`

`=> x^2 = 0.74 xx 10^(-12)`

`x = 0.86 xx 10^(-6)`

`["OH"^-] = 0.86 xx 10^(-6)`

`therefore ["H"^+] = "K"_"w"/(0.86 xx 10^(-6))`

`= 10^(-14)/(0.86 xx 10^(-6))`

`["H"^+] = 1.162 xx 10^(-8)`

`"pH" = - log["H"^+]`

= 7.94