We are asked to calculate the solubility of cobalt(II) hydroxide, Co(OH)₂, in pure water and in a 0.0080 M CoCl₂ solution at 25°C. Given that the solubility product constant $K_{sp}$ of Co(OH)₂ is $5.92 \times 10^{-15}$, we will compute the solubility in both conditions.

Part 1: Solubility of Co(OH)₂ in pure water

For Co(OH)₂, the dissociation reaction is:

$\text{Co(OH)}_2 (s) \rightleftharpoons \text{Co}^{2+} (aq) + 2\text{OH}^- (aq)$

The solubility product expression is:

[ K_{sp} = [\text{Co}^{2+}] [\text{OH}^-]^2 ]

Let $s$ be the molar solubility of Co(OH)₂ in pure water. In pure water, we assume:

[ [\text{Co}^{2+}] = s \quad \text{and} \quad [\text{OH}^-] = 2s ]

Thus, the $K_{sp}$ expression becomes:

$K_{sp} = s(2s)^2 = 4s^3$

Now, substitute the given value of $K_{sp} = 5.92 \times 10^{-15}$:

$5.92 \times 10^{-15} = 4s^3$

Solving for $s$:

$s^3 = \frac{5.92 \times 10^{-15}}{4} = 1.48 \times 10^{-15}$

$s = \sqrt[3]{1.48 \times 10^{-15}} = 1.14 \times 10^{-5} \, \text{M}$

Thus, the molar solubility of Co(OH)₂ in pure water is $1.14 \times 10^{-5} \, \text{M}$.

To convert this to grams per liter (g/L), we use the molar mass of Co(OH)₂:

$\text{Molar mass of Co(OH)}_2 = 58.93 + 2(16.00 + 1.01) = 93.95 \, \text{g/mol}$

$\text{Solubility in g/L} = 1.14 \times 10^{-5} \, \text{M} \times 93.95 \, \text{g/mol} = 1.07 \times 10^{-3} \, \text{g/L}$

Part 2: Solubility of Co(OH)₂ in 0.0080 M CoCl₂ solution

In a 0.0080 M CoCl₂ solution, the concentration of $\text{Co}^{2+}$ is no longer $s$, but rather $0.0080 \, \text{M} + s$. Since $s$ is much smaller compared to 0.0080 M, we can approximate $[\text{Co}^{2+}] \approx 0.0080 \, \text{M}$.

The $K_{sp}$ expression is still:

[ K_{sp} = [\text{Co}^{2+}] [\text{OH}^-]^2 ]

Substituting known values:

$5.92 \times 10^{-15} = (0.0080) [\text{OH}^-]^2$

Solving for $[\text{OH}^-]$:

$[\text{OH}^-]^2 = \frac{5.92 \times 10^{-15}}{0.0080} = 7.40 \times 10^{-13}$

$[\text{OH}^-] = \sqrt{7.40 \times 10^{-13}} = 8.60 \times 10^{-7} \, \text{M}$

Since $[\text{OH}^-] = 2s$, we have:

$2s = 8.60 \times 10^{-7} \quad \Rightarrow \quad s = \frac{8.60 \times 10^{-7}}{2} = 4.30 \times 10^{-7} \, \text{M}$

Now, convert this molar solubility to g/L:

$\text{Solubility in g/L} = 4.30 \times 10^{-7} \, \text{M} \times 93.95 \, \text{g/mol} = 4.04 \times 10^{-5} \, \text{g/L}$

Final Answers:

• Solubility of Co(OH)₂ in pure water: $1.1 \times 10^{-3} \, \text{g/L}$
• Solubility of Co(OH)₂ in 0.0080 M CoCl₂ solution: $4.0 \times 10^{-5} \, \text{g/L}$

Would you like further details on any part of the calculations?

Here are some related questions to explore:

1. What is the relationship between solubility and molar solubility?
2. How does the common ion effect reduce solubility?
3. What factors affect the solubility product constant $K_{sp}$?
4. How can we calculate pH from the solubility of a hydroxide salt?
5. What is the significance of the third cube root in solubility calculations?

Tip: In solutions containing common ions, solubility is suppressed due to the common ion effect, where the excess of one ion shifts the equilibrium to lower solubility.