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

`[HSaC] = ("Mole")/("Litre") = (4 xx 10^(-4))/(200//1000) = 2 xx 10^(-3)M` M
The dissociation of `HSaC` takes place in presence of `[H^(o+)] = 10^(-3)M`
`{:(,HSaChArr,H^(o+)+,SaC^(Theta),),(["Concentration before dissociation"],2xx10^(-3),10^(-3),0,):}`
In presence of `H^(o+)` the dissociation of `HSaC` in almost negligible because of the common ion effect. Thus, at equilibrium
`[HSaC] = 2 xx 10^(-3), [H^(o+)] = 10^(-3)`
`:. K_(a) = ([H^(o+)][SaC^(Theta)])/([HSaC]) :. 2 xx 10^(-12) = ([10^(-3)][SaC^(-)])/(2xx10^(-3))`
`:. [SaC^(Theta)] = 4 xx 10^(-12)M`