Calculate the solubility at 25 celsius of Co(OH)2 in pure water and in a 0.0080M CoCl2 solution. ksp of Co(OH)2 is 5.92×10^-15
Round both of your answers to 2 significant digits.
solubility in pure water:	 g/l
solubility in 0.0080 M CoCl2 solution:	g/l

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.

Calculate the solubility at 25 celsius of Co(OH)2 in pure water and in a 0.0080M CoCl2 solution. Ksp of Co(OH)2 is 5.92×10^-15. Round both of your answers to 2 significant digits. Solubility in pure water: g/l, solubility in 0.0080 M CoCl2 solution: g/l

This solution covers how to calculate the solubility of cobalt(II) hydroxide (Co(OH)₂) in pure water and a 0.0080 M CoCl₂ solution at 25°C. It explores chemical equilibrium concepts such as Ksp and the common ion effect, with step-by-step calculations involving algebra, cube roots, and molar mass conversion. Suitable for students in general chemistry or undergraduate-level courses.

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