Calculate the solubility of Co(OH)2, in g/L, in solutions that have been buffered to the following pHs. Ksp=1.6*10^-15.

a. 7.00
b. 10.00
c. 4.00

a. 14.9g/L

b. 1.49x10⁻⁶

c. 1.49x10⁷

Explanation:

You can write the buffer Ksp of Co(OH)₂ as follows:

Co(OH)₂(s) ⇄ Co²⁺ + 2OH⁻

Ksp = 1.6x10⁻¹⁵ = [Co²⁺] [OH⁻]²

To have buffered the solutions means  [OH⁻] is fixed. From the equilibrium of water we can relate  [OH⁻] with pH as follows:

[OH⁻] = 10^[14-pH]

With [OH⁻] and Ksp we can solve for [Co²⁺]. Its concentration is equal to solubility (That is the amount of Co(OH)₂ that can be dissolved).

[Co²⁺] is in mol/L. With molar mass of Co(OH)₂ -92.948g/mol-, We can obtain, in the end, its solubility in g/L.

-Molar concentration of [Co²⁺] and solubility:

a. [OH⁻] = 10^[14-7.00] = 1x10⁻⁷

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁷]²

[Co²⁺] = 0.16mol / L = Solubility.

In g/L = 0.16mol / L ₓ(92.948g/mol) =

14.9g/L

b. [OH⁻] = 10^[14-10.00] = 1x10⁻⁴

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁴]²

[Co²⁺] = 1.6x10⁻⁸mol / L = Solubility.

In g/L = 1.6x10⁻⁸mol / L ₓ(92.948g/mol) =

1.49x10⁻⁶g/L

c.[OH⁻] = 10^[14-4.00] = 1x10⁻¹⁰

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻¹⁰]²

[Co²⁺] = 1.6x10⁵mol / L = Solubility.

In g/L = 1.6x10⁵mol / L ₓ(92.948g/mol) =

1.49x10⁷g/L

As you can see, and as general rule, all hydroxides are solubles in acids.

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

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