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

Options

A. `1 xx 10^(-4)`
B. `1 xx 10^(-5)`
C. `1 xx 10^(-6)`
D. `1 xx 10^(-3)`

Answer

A

Explanation

a) Rate with respect to weak acid.
`R_(1) = k[H^(+)]_("weak acid")`
and rate with respect to strong acid.
`R_(2) = k[H^(+)]_("strong acid")`
`R_(1)/R_(2) = ((kH^(+))_("weak acid"))/(kH^(+))_("strong acid")=1/100`
`[H^(+)]_("weak field")=1/10=0.01M`
`underset(1-0.01~~1)HA ltimplies underset(0.01)H^(+) + underset(0.01)A^(-1)`
`K_(a) = ([H^(+)][A^(-)])/([HA])=(0.01 xx 0.01)/(1) = 1 xx 10^(-4)`