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

Question

Mateo Aguilar · Accepted Answer

As discussed below.

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).

Cooper Adams · Answer

undefined The Ksp of cesium sulfate (Cs2SO4) is approximately 3.16 x 10^-31.

Kai Bowman · Answer

undefined The $K_{sp}$ (solubility product constant) of cesium sulfate ($Cs_2SO_4$) is approximately $2.7 \times 10^{-6}$ at 25°C.