Question

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

Answer 1

As discussed below.

Explanation:

The solubility product constant `K_"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_"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):

`"A"(s)⇌c"C"(aq)+d"D"(aq)` ..........(1)

And solubility equilibrium general expression
`K_c=(["C"]^c["D"]^d)/(["A"])`
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_cxx["A"]=["C"]^c["D"]^d`
Writing LHS as solubility product constant `K_"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
`"Cs"_2"SO"_4(s)⇌2"Cs"^(2+)(aq)+"SO"_4^(2-)(aq)` .....(3)

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

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

`"Cs"_2"SO"_4=` `167" at "0^@C,` `173" at "15^@C,` `179" at "20^@C.`

To calculate value of `K_"sp"` we need molar solubility of at least one of the three in equation (3).