Question

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 1

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)

K_sp = [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