InterviewSolution
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InterviewSolution
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The solubility of calcium sulphate is `0.67 g L^(-1)` . The value of `K_(sp)` for calcium sulphate will beA. `1.7xx10^(-6)`B. `3.5xx10^(-4)`C. `2.4xx10^(-5)`D. `9.3xx10^(-8)` |
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Answer» Correct Answer - C 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)` |
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