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

1 Answer
Jun 13, 2016

Answer:

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).