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.

Explanation

We know,
`[HSaC] = ("mole")/("litre")= (4xx10^(-4))/(200//1000)=2xx10^(-3)M`
The dissociation of HSaC takes place in presence of `[H^(+)]=10^(-3)`
`{:(,HSaChArr,H^(+)+,SaC^(-)),("Conc. Before dissociation", 2xx10^(-3),10^(-3),0):}`
In pressence of `H^(+)`, the dissociation of HSaC is almost negligible because of common ion effect. Thus, at equilibrium.
`[HSaC]=2xx10^(-3),[H^(+)]=10^(-3)`
`K_(a)=([H^(+)][SaC^(-)])/([HSaC])`
`:. 2xx10^(-12)=([10^(-3)][SaC^(-)])/([2xx10^(-3)])`
`:. [SaC^(-)]=4xx10^(-12)M`