# What is the K_(sp) of cesium sulfate?

Jun 13, 2016

As discussed below.

#### Explanation:

The solubility product constant ${K}_{\text{sp}}$​, is defined for equilibrium between a solids and its respective ions in its aqueous solution. It represents the level at which a solute dissolves in solution. Higher the ${K}_{\text{sp}}$​ value means more soluble a substance is.

This constant is used to describe saturated solutions of ionic compounds of relatively low solubility. A saturated solution is in a state of dynamic equilibrium between the dissolved, dissociated, ionic compound and the undissolved solid.
Consider the general dissolution reaction below (in aqueous solutions):

$\text{A"(s)⇌c"C"(aq)+d"D} \left(a q\right)$ ..........(1)

And solubility equilibrium general expression
${K}_{c} = \left(\left[\text{C"]^c["D"]^d)/(["A}\right]\right)$
Where terms in the numerator represent concentrations of respective ions raised to power equal to the number of each ion in the solution and denominator is a value representing an amount of solid solute in moles in a litre. We see that the denominator is a constant, therefore, rewriting we get the following equation:
${K}_{c} \times {\left[\text{A"]=["C"]^c["D}\right]}^{d}$
Writing LHS as solubility product constant ${K}_{\text{sp}}$, we get

K_"sp"=["C"]^c["D"]^d ....(2)

At equilibrium in given conditions an expression like (1) above for cesium sulfate can be written as
${\text{Cs"_2"SO"_4(s)⇌2"Cs"^(2+)(aq)+"SO}}_{4}^{2 -} \left(a q\right)$ .....(3)

:. K_"sp"=["Cs"^(2+)]^2["SO"_4^(2-)] ......(4)

From tables solubility of Cesium sulfate in g/100g of water is

${\text{Cs"_2"SO}}_{4} =$ $167 \text{ at } {0}^{\circ} C ,$ $173 \text{ at } {15}^{\circ} C ,$ $179 \text{ at } {20}^{\circ} C .$

To calculate value of ${K}_{\text{sp}}$ we need molar solubility of at least one of the three in equation (3).