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Question: How to calculate \( ~Ksp \) for \( CuS{{O}_{4}} \) and \( C{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}...

How to calculate  Ksp~Ksp for CuSO4CuS{{O}_{4}} and Ce2(SO4)3C{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}} ?

Explanation

Solution

Hint : Solubility product is the product of the ionic concentrations or activities of an electrolyte and the molar solubility is the number of moles of salts that can be dissolved in one litre of solution to form a saturated solution and then derive the required equation which gives the answer.

Complete Step By Step Answer:
In the classes of physical chemistry, we have studied about the concept of molar solubility and also about the solubility product and some of the general definitions relating to it. Let us now see how the solubility product can be calculated from the molar solubility.
Firstly let us understand what solubility product is and what the molar solubility states.
Solubility product is the maximum product of the ionic concentrations or the activities of an electrolyte that at one temperature will be able to continue in equilibrium with dissolved phase or in simple word say can say that  Ksp~Ksp is equilibrium constant for a solid substance dissolving in aqueous solution.
Here, we have CuSO4(s)⇌Cu2++SO42−CuS{{O}_{4(s)}}\rightleftharpoons C{{u}^{2+}}+SO_{4}^{2-}
Thus, Ksp=[Cu2+][SO42−]Ksp=[C{{u}^{2+}}][SO_{4}^{2-}]
We are quoted a solubility of 24.06g⋅L−1 24.06g\cdot {{L}^{-1}}~ with respect to copper sulfate. Molar solubility=(24.06g159.61g⋅mol−1)1L=0.151mol⋅L−1Molar\text{ }solubility=\dfrac{\left( \dfrac{24.06g}{159.61g\cdot mo{{l}^{-1}}} \right)}{1L}=0.151mol\cdot {{L}^{-1}} and thus we have [Cu2+]=[SO42−]=0.151mol⋅L1[C{{u}^{2+}}]=[SO_{4}^{2-}]=0.151mol\cdot {{L}^{1}}
Thus we get KspKsp can be rewritten as; Ksp=(0.151mol⋅L−1)2=2.28×10−2Ksp={{(0.151mol\cdot {{L}^{-1}})}^{2}}=2.28\times {{10}^{-2}}
Now for Ce2(SO4)3(s)⇌2Ce3++3SO42−C{{e}_{2}}{{(S{{O}_{4}})}_{3(s)}}\rightleftharpoons 2C{{e}^{3+}}+3SO_{4}^{2-}
Thus Ksp=[Ce3+]2[SO42−]3Ksp={{\left[ C{{e}^{3+}} \right]}^{2}}{{\left[ SO_{4}^{2-} \right]}^{3}}
Molar solubility=(6.22g322.24g⋅mol−1)1L=0.0187mol⋅L−1Molar\text{ }solubility=\dfrac{\left( \dfrac{6.22g}{322.24g\cdot mo{{l}^{-1}}} \right)}{1L}=0.0187mol\cdot {{L}^{-1}} and thus we have Ksp=[Ce3+]2[SO42−]3=(2×0.0187)2(3×0.0187)3=2.469×10−7Ksp={{\left[ C{{e}^{3+}} \right]}^{2}}{{\left[ SO_{4}^{2-} \right]}^{3}}={{\left( 2\times 0.0187 \right)}^{2}}{{\left( 3\times 0.0187 \right)}^{3}}=2.469\times {{10}^{-7}}
⇒Ksp=2.469×10−7\Rightarrow Ksp=2.469\times {{10}^{-7}}

Note :
Note that the molar solubility is related to the solubility product constant KspKsp as the higher the value of the solubility product, the more soluble the compound will be and this says that they are directly related to each other.