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The ionization constant of chloroacetic acid is 1.35xx10^(-3). What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution ? |
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Answer» Solution :`ClCH_(2)CO OH hArr ClCH_(2)CO O^(-)+ H^(+),K_(a)=([ClCH_(2)CO O^(-)][H^(+)])/([ClCH_(2)CO OH])=([H^(+)]^(2))/(C)` `[H^(+)]=sqrt(K_(a)xxc)=sqrt(1.35xx10^(-3)xx0.1)=1.16xx10^(-2)M` `pH=-log (1.16xx10^(-2))=2-0.06=1.94` Sodium salt of chloroacetic acid is a salt of strong base and weak acid . Hence, `pH = - (1)/(2) [ log K_(W)+logK_(a)-logc]` `:. pH = -(1)/(2) [log 10^(-14)+log 1.35xx10^(-3)-log 0.1]` `=-(1)/(2) [-14+(-3+0.1303)-(-1)]=7+1.44-0.5=7.94` |
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