Questions 9 & 10 refer to the solubility of cadmium hydroxide, Cd(OH)_2: Cd(OH)_

Question

Questions 9 & 10 refer to the solubility of cadmium hydroxide, Cd(OH)_2: Cd(OH)_2(s) Cd^2+(aq) + 2OH^-(aq) The K_sp for Cd(OH)_2 is 5.9 times 10^-15 at 25 degree C. What is the solubility of Cd(OH)_2 in mol L^-1? a) 7.7 times 10^-8 b) 5.4 times 10^-8 c) 1.1 times 10^-5 d) 5.9 times 10^-15 e) 1.8 times 10^-5 If 100. mL of 2.0 times 10^-5 M Cd(NO_3)_2 is added to 100. mL of a solution of 2.0 times 10^-5 M KOH, which statement is correct? a) The ionic product is 1.0 times 10^-15 and Cd(OH)_2 (s) does not precipitate. b) The ionic product is 1.0 times 10^--15 and Cd(OH)_2(s) precipitates c) The ionic product is 1.0 times 10^-10 and Cd(OH)_2(s) precipitates d) The ionic product is 1.0 times 10^-10 and Cd(OH)_2(s) does not precipitate e) none of the above

Explanation / Answer

9. let x amount of Cd(OH)2 is in solution then,

Ksp = [Cd2+][OH-]^2

5.9 x 10^-15 = (x)(2x)^2

molar solubility (x) = 1.1 x 10^-5 mol/L

Answer : c) 1.1 x 10^-5 mol/L

10. Qsp = [Cd2+][OH-]^2

              = (2 x 10^-5 M x 0.1 L/0.2 L)(2 x 10^-5 M x 0.1 L/0.2 L)^2

              = 1 x 10^-15

As Ksp > Qsp, no precipitation would occur.

True statement would thus be,

a) The ionic product is 1.0 x 10^-15 and Cd(OH)2 does not precipitate.

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