Saccharin `(K_(a)=2xx10^(-12))` is a weak acid represented by formula HSaC. A `4xx10^(-4)` mole amount of saccharin is dissolved in 200 `cm^(3)` water of pH 3. Assuming no change in volume. Calculate the soncentration of `SaC^(-)` ions in the resulting solution at equilibrium.

Options

A. `2 xx 10^(-12)M`
B. `3 xx 10^(-12)M`
C. `4 xx 10^(-12)M`
D. `5 xx 10^(-12)M`

Answer

C

Explanation

`[HSaC]= (4 xx 10^(-4))/(200//1000) =2 xx 10^(-3)M`
The dissociation of HSaC takes places in the pres- ence of `[H^+]= 10^(-3)` `"conc. before dissociation" 2xx 10^(-3)" "10^(-3)" "0`
In presence of `H^+` the dissociation of HSaC is al- most negligible because of common ion effect. Thus at equilibrium `[HSac]=2 xx 10^(-3), H^(+)=10^(-3)`
`K_(a)=([H^(+)][SaC^(-)])/([HSaC])`
`therefore 2 xx 10^(-12)= ([10^(-3)][SaC^(-)])/([2 xx 10^(-3)])`
`therefore [SaC^(-)]=4 xx 10^(-12)M`