When 100.0 mL of 2.00 M Ce(NO3)3 is added to 100.00mL of 3.00M KIO3, a precipitate of Ce(IO3)3 (s) forms. Calculate the equilibrium concentrations of Ce3+ and IO- in this solution. Ksp Ce(IO3)3=3.2x10^-10
Answer
[Ce3+] = 0.50029M
[IO3-] = 0.00086M
Explanation
Initial concentration of Ce3+ = 2.00M /(200ml/100ml) = 1.00M
Initial concentration of IO3- = 3.00M/( 200ml/100ml) = 1.50M
consider the complete formation of Ce(IO3)3
Ce3+(aq) + 3IO3-(aq) -------> Ce(IO3)3(s)
3mole of IO3- reacts with 1mole of Ce3+
1.50M of IO3- react with 0.50M of Ce3+
after completion of reaction
Concentration of Ce3+ = 0.50M
Concentration of IO3- = 0
Now , consider the solubility equilibrium of Ce(IO3)3
Ce(IO3)3(s) <-------> Ce3+(aq) + 3IO3-(aq)
Ksp = [Ce3+] [IO3-]3 = 3.2 ×10-10
Initial concentration
[Ce3+] = 0.50
[IO3-] = 0
change in concentration
[Ce3+] = + x
[IO3-] = + 3x
Equilibrium concentration
[Ce3+] = 0.50 + x
[IO3-] = 3x
so,
(0.50 + x) ( 3x)3 = 3.2 ×10-10
solving for x
x = 0.0002872M
Therefore,
at equilibrium
[Ce3+] = 0.50 + ( 0.0002872) = 0.50029M
[IO3-] = 3× (0.0002872) = 0.00086M
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