The solubility of calcium sulphate is `0.67 g L^(-1)` . The value of `K_(sp)` for calcium sulphate will be

Options

A. `1.7xx10^(-6)`
B. `3.5xx10^(-4)`
C. `2.4xx10^(-5)`
D. `9.3xx10^(-8)`

Answer

C

Explanation

Calculate the number of moles of `CaSO_(4)` dissolved in 1 L of solution (formula mass of `CaSO_(4)` is `40+32+4xx16= 136 u)`
`(0.67g CaSO_(4))/(1L "soln")xx(1mol CaSO_(4))/(136g CaSO_(4))= 4.9xx10^(-3)mol^(-1)`
Form the solubility equilibrium
`CaSO_(4)(s)hArrCa^(2+)(aq.)+SO_(4)^(2-)(aq.)`
we observe that each mole of `CaSO_(4)` yields 1 mol of `Ca^(2+)` and 1 mol of `SO_(4)^(2)`. Thus, at equilibrium (saturated solution).
`C_(Ca^(2+))=4.9xx10^(-3)mol L^(-1)`
`C_(SO_(4)^(2-))= 4.9xx10^(-3)mol L^(-1)`
`K_(sp)=C_(Ca^(2+))C_(SO_(4)^(2-))`
`= (4.9xx10^(-3))(4.9xx10^(-3))`
`= 2.4xx10^(-5)`