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Calculate the pH at equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M NaOH solution . `K_(a) ` for acetic acid `= 1.9xx10^(-5)`

Answer» At the equivalence point, `CH_(3)CO ON a` is formed asn its concentration `=(0.1)/(2) M = 0.05 M`.
It is a slat of weak acid and strong base. The formula for finding the pH of such a salt is
`pH = - (1)/(2) [log K_(w) + log K_(a) - log c]`
`:. pH = - (1)/(2) [log 10^(-14)+log (1.9xx10^(-5))-log(5xx10^(-2))]`
`=-(1)/(2) [-14 + (-5+0.2788)-(-2+0.6990)]=(1)/(2) (14+5-0.2788-2+0.6990)=(17.42)/(2) = 8.71`


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