Answer:
a. 14.9g/L
b. 1.49x10⁻⁶
c. 1.49x10⁷
Step-by-step 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.