The solubilityproduct constant for Ce10 is 32 1010 What is

The solubility-product constant for Ce(10), is 3.2 × 10-10 What is the Ce3+ concentration in a solution prepared by mixing 40.00 mL of 0.0440 M Ce3+ with 40.00 mL of a. water? b. 0.0440 M 103-? c. 0.240 M 103_? d. 0.0490 M IO3?

Solution

a)

Initial Concentration of Ce³⁺ is diluted for 2 time

Therefore,

Final Concentration of Ce³⁺ = 0.0440M/2 = 0.0220M

b)

Ce(IO3)3(s) <-----> Ce³⁺(aq) + 3IO3^-(aq)

Ksp = [Ce³⁺] [ IO3^-]³ = 3.2×10^-10

[IO3^-] = 0.0440M/2 = 0.0220M

Therefore

[Ce³⁺] × (0.0220M)³ = 3.2×10^-10

[Ce³⁺] = 3.2×10^-10 / 1.0648×10^-5

[Ce³⁺] = 3.01×10^-5M

c)

[IO3^-] = 0.240M/2 = 0.120M

[Ce³⁺] × (0.120M)³ = 3.2×10^-10

[Ce³⁺]= 3.2×10^-10/1.728 ×10^-3

[Ce³⁺] = 1.85×10^-7M

d)

[IO3^-] = 0.0490M / 2 = 0.0245M

[Ce³⁺]×(0.0245)³= 3.2×10^-10

[Ce³⁺]= 3.2×10^-10/1.471×10^-5

   [Ce³⁺] = 2.18×10^-5M