Question

1. Calculate the molar solubility of Cr(OH), in a solution with a pH of 11.5 knowing that Kop(Cr(OH)2) = 6.7 x 10"

calculate the molar solubility of Cr(OH)3 in a solution with a pH of 11.5 knowing that Ksp(Cr(OH)3) = 6.7 x 10^(-31)

Answer 1

1)

use:

pH = -log [H+]

11.5 = -log [H+]

[H+] = 3.162*10^-12 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(3.162*10^-12)

[OH-] = 3.162*10^-3 M

At equilibrium:

Cr(OH)3 <----> Cr3+ + 3 OH-

   s 3.162*10^-3

Ksp = [Cr3+][OH-]^3

6.7*10^-31=(s)*(3.162*10^-3)^3

6.7*10^-31= (s) * 3.161*10^-8

s = 2.119*10^-23 M

Answer: 2.12*10^-23 M

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