Question

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid `(HA,1M)` is `1//100th` of that of a strong acid `(HX, 1M),` at `25^(@)C`. The `K_(a)(HA)` is
A. `1xx10^(-3)`
B. `1xx10^(-4)`
C. `1xx10^(-5)`
D. `1xx10^(-6)`

Answer 1

Correct Answer - B
The hydrolysis of an ester is catalyzed by `H^(+)` ions: `CH_(3)COOH_(3)(aq.)+H_(2)O overset(H^(+))hArr CH_(3)COOH(aq.)+CH_(3)OH(aq.)`
Thus, the rate of hydrolysis is direactly related to `C_(H^(-)`. Since rate in weak acid `=(1)/(100)xx` (rate in strong acid)
We have `[C_(H^(+))]_("Weak acid")=(1)/(100)[C_(H^(+))]_("Strong acid")`
`=(1)/(100)(1M)`
`=10^(-2)M`
`{:(,HA(aq.)hArrH^(+)(aq.)+A^(-)(aq.)),("Initial (M)"," C 0 0"),("Change (M)",-Calpha " "+C alpha " "+C alpha),("Equilibrium (M)",bar(C- C alpha " "C alpha " "C alpha)):}`
Thus, `C_(H^(+))=Calpha=10^(-2)M`
`alpha=10^(-2)M//C=10^(-2)M`
`K_(a)=(C_(H^(+)C_(A)^(-)))/(C_(HA))=((Calpha)(Calpha))/(C(1-alpha))`
`=Calpha^(2)`(Neglecting `alpha` with respect to I)
`=(1)(10^(-2))=10^(-4)`