Question

Part 1: Calculate the solubility (in M) of cobalt(II) hydroxide, Co(OH)2(s) in H2O. Ksp = 2.80×10-16 at a specific temperature

Part 2; What is the pH of this solution of cobalt(II) hydroxide?

Answer 1

part1

Co(OH)2(s) ---------------> Ca^2+ (aq) + 2OH^- (aq)

                                           x                  2x

Ksp    = [Co^2+][OH^-]^2

2.8*10^-16   = x*(2x)^2

2.8*10^-16   = 4x^3

x^3               = 7*10^-17

x   = 4.12*10^-6

solubility of Co(OH)2 in water = 4.12*10^-6M

part-b

Co(OH)2(s) ---------------> Ca^2+ (aq) + 2OH^- (aq)

                                           x                  2x

Ksp    = [Co^2+][OH^-]^2

2.8*10^-16   = x*(2x)^2

2.8*10^-16   = 4x^3

x^3               = 7*10^-17

x   = 4.12*10^-6

[OH^-]   = 2x = 2*4.12*10^-6

                        = 8.24*10^-6M

POH   = -log[OH^-]

            = -log8.24*10^-6

              =5.084

PH   = 14-POH

          = 14-5.084

          = 8.916>>>>answer