The solubility product ofa sparingly soluable salt AB at room temperature is `1.21 xx 10^(-6).` Its molar solubility is

Options

A. `1.21 xx 10^(-6)`
B. `1.21 xx 10^(-3)`
C. `1.1 xx 10^(-4)`
D. `1.1 xx 10^(-3)`

Answer

D

Explanation

For a saturated solution of a sparingly soluable salt AB, the following solubility equilibrium would exist AB (Saturated salt solution) `hArr A^(+) (aq) + B^(-)(aq)`
If the molar solubility of the salt S, then `[A^(+)]` = S mol `dm^(-3), [B^(-)]` = S mol `dm^(-3)`. Therefore, solubility product is given by
`K_(sp) = [A^(+)][B^(-)] = (S)(S) = S^(2)`
Given that `K_(sp) = 1.21 xx 10^(-6)`, so the molar solublity is, `S = sqrt(K_(sp)) = sqrt(1.21 xx 10^(-6)) = 1.1 xx 10^(-3)`