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Calculate K_{sp} values of the following, for which a reference book lists the indicated solubilities: (a) Ce(IO_{3})_{4}, solubility 1.8×10^{–4} mol/L; (b) Hg_{2}SO_{4}, 8.9×10^{–4} mol/L; and (c) barium chromate, 0.0010 g BaCrO_{4}/100 mL H_{2}O.
Strategy: write the chemical reaction, write the mass action expression, set up a table of concentrations, convert the given solubility to molarity if necessary, and then use this value to find K_{sp}.
(a)
Ce(IO_{3})_{4}(s) 
Ce^{4+}(aq) 
+ 
4 IO_{3}^{–}(aq) 

K_{sp} = [Ce^{4+}]_{e}[IO_{3}^{–}]_{e}^{4} 

Initial 
0 
0 

Change 
+ x 
+ 4x 

Equilibrium 
x 
4x 
Molar solubility of Ce(IO_{3})_{4} = x = 1.8×10^{–4} mol/L
[Ce^{4+}]_{e} = x = 1.8×10^{–4} mol/L
[IO_{3}^{–}]_{e} = 4x = 4(1.8×10^{–4} mol/L) = 7.2×10^{–4} mol/L
K_{sp} = [Ce^{4+}]_{e}[IO_{3}^{–}]_{e}^{4} = [1.8×10^{–4} mol/L][7.2×10^{–4} mol/L]^{4} = 4.8×10^{–17}
(b)
Hg_{2}SO_{4}(s) 
Hg_{2}^{2+}(aq) 
+ 
SO_{4}^{2–}(aq) 

K_{sp} = [Hg_{2}^{2+}]_{e}[SO_{4}^{2–}]_{e} 

Initial 
0 
0 

Change 
+ x 
+ x 

Equilibrium 
x 
x 
Molar solubility of Hg_{2}SO_{4} = x = 8.9×10^{–4} mol/L
[Hg_{2}^{2+}]_{e} = x = 8.9×10^{–4} mol/L
[SO_{4}^{2–}]_{e} = x = 8.9×10^{–4} mol/L
K_{sp} = [Hg_{2}^{2+}]_{e}[SO_{4}^{2–}]_{e} = [8.9×10^{–4} mol/L][8.9×10^{–4} mol/L] = 7.9×10^{–7}
(c)








Initial 



Change 



Equilbrium 


The variable x must be found from the solubility given, after converting to molarity.
BaCrO_{4} = 137.3 + 52.0 + 4(16.0) = 253.3 g/mol
0.0010 g/100 mL = 0.010 g/L = 0.010 g/(253.3 g/mol)/L = 3.9×10^{–5} M
Since the stoichiometry between barium chromate and the ions is 1:1, x = 3.9×10^{–5} M.
Now this value can be used in the mass action expression for K_{sp} to give:
K_{sp} = [x][x] = [3.9×10^{–5}] [3.9×10^{–5}] = 1.6×10^{–9}