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# What is the ${K_{sp}}$ of cesium Sulfate? Verified
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Hint: The equilibrium between a solid and its respective ions in its aqueous solution is described by the solubility product constant ${K_{sp}}$ . It denotes the concentration at which a solute dissolves in water. A higher ${K_{sp}}$ value indicates that a material is more soluble.

This constant is used to characterize saturated solutions of low-solubility ionic compounds. The dissolved, dissociated ionic compound and the undissolved solid are in a state of dynamic equilibrium in a saturated solution. Consider the following (in aqueous solutions) general dissolution reaction:
${\text{A(s)}} \rightleftharpoons {\text{cC(aq) + dD(aq) - - - - - - (1)}}$
Also, the general expression for solubility equilibrium:
${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{[A]}}$
Where the numerator represents the quantities of respective ions raised to the power of the number of each ion in the solution, and the denominator represents the volume of solid solute in moles per litre. We can see that the denominator is a constant, so we can rewrite the equation as follows:
${K_c} \times [A] = {[C]^c}{[D]^d}$
When we write LHS as ${K_{sp}}$ , the solubility product constant, we get
${K_{sp}} = {[C]^c}{[D]^d} - - - - - (2)$
For cesium sulphate, at equilibrium in given conditions, an expression like (1) can be written as
$C{s_2}S{O_4}(s) \rightleftharpoons 2C{s^{2 + }}(aq) + S{O^{2 - }}_4(aq) - - - - (3) \\ \therefore {K_{sp}} = {[C{s^{2 + }}]^2}[S{O_4}^{2 - }] - - - - (4) \\$
The solubility of Cesium sulphate in g/ $100$ g of water is calculated from tables.
$C{s_2}S{O_4} = 167\,at\,{0^0}C,\,173\,at\,{15^0}C,\,179\,at\,{20^0}C$
We need the molar solubility of at least one of the three in equation(3) to determine the value of ${K_{sp}}$ .

Note:
The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, ${K_{sp}}$ . It denotes the concentration at which a solute dissolves in water.