What is the molar solubility of zinc hydroxide in a $$ 0.0010 \mathrm{M} \mathrm{ZnCl}_{2} $$ solution? $$ \left(\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Zn}(\mathrm{OH})_{2} ight)=3.0 imes 10^{-17} ight) $$ $$ \begin{array}{l}8.7 imes 10^{-8} \mathrm{M} \ 2.0 imes 10^{-5} \mathrm{M} \ 3.0 imes 10^{-17} \mathrm{M} \ 2.0 imes 10^{-6} \mathrm{M} \ 4.3 imes 10^{-8} \mathrm{M} \ \ ext { Save for Later }\end{array} $$

Question

What is the molar solubility of zinc hydroxide in a $$ 0.0010 \mathrm{M} \mathrm{ZnCl}_{2} $$ solution? $$ \left(\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Zn}(\mathrm{OH})_{2}ight)=3.0 	imes 10^{-17}ight) $$ $$ \begin{array}{l}8.7 	imes 10^{-8} \mathrm{M} \ 2.0 	imes 10^{-5} \mathrm{M} \ 3.0 	imes 10^{-17} \mathrm{M} \ 2.0 	imes 10^{-6} \mathrm{M} \ 4.3 	imes 10^{-8} \mathrm{M} \ \ 	ext { Save for Later }\end{array} $$

UpStudy AI · Accepted Answer

To find the molar solubility of zinc hydroxide ($$ \text{Zn(OH)}_2 $$) in a $$ 0.0010 \, \text{M} \, \text{ZnCl}_2 $$ solution, we can follow these steps:

1. **Write the dissociation equation for zinc hydroxide:**
   $$
   \text{Zn(OH)}_2 (s) \rightleftharpoons \text{Zn}^{2+} (aq) + 2 \text{OH}^- (aq)
   $$

2. **Define the solubility product constant ($$ K_{sp} $$):**
   The solubility product expression for zinc hydroxide is given by:
   $$
   K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2
   $$
   Given that $$ K_{sp} = 3.0 \times 10^{-17} $$.

3. **Determine the concentration of $$ \text{Zn}^{2+} $$:**
   In the $$ 0.0010 \, \text{M} \, \text{ZnCl}_2 $$ solution, the concentration of $$ \text{Zn}^{2+} $$ is already $$ 0.0010 \, \text{M} $$.

4. **Let $$ s $$ be the molar solubility of $$ \text{Zn(OH)}_2 $$:**
   When $$ \text{Zn(OH)}_2 $$ dissolves, it produces $$ s $$ moles of $$ \text{Zn}^{2+} $$ and $$ 2s $$ moles of $$ \text{OH}^- $$. Therefore, the concentrations will be:
   - $$ [\text{Zn}^{2+}] = 0.0010 + s $$
   - $$ [\text{OH}^-] = 2s $$

5. **Substituting into the $$ K_{sp} $$ expression:**
   Since $$ s $$ is expected to be very small compared to $$ 0.0010 \, \text{M} $$, we can approximate:
   $$
   [\text{Zn}^{2+}] \approx 0.0010
   $$
   Thus, the $$ K_{sp} $$ expression becomes:
   $$
   K_{sp} = (0.0010)(2s)^2
   $$
   Simplifying this gives:
   $$
   K_{sp} = (0.0010)(4s^2) = 4.0 \times 10^{-3} s^2
   $$

6. **Setting up the equation:**
   $$
   3.0 \times 10^{-17} = 4.0 \times 10^{-3} s^2
   $$

7. **Solving for $$ s $$:**
   $$
   s^2 = \frac{3.0 \times 10^{-17}}{4.0 \times 10^{-3}}
   $$
   $$
   s^2 = 7.5 \times 10^{-15}
   $$
   $$
   s = \sqrt{7.5 \times 10^{-15}}
   $$

Now, let's calculate $$ s $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{7.5e-15}$$
- step1: Factor the expression:
  $$\sqrt{\frac{15}{2}\left(e-2\right)}$$
- step2: Use the properties of radicals:
  $$\sqrt{\frac{15}{2}}\times \sqrt{e-2}$$
- step3: Evaluate:
  $$\frac{\sqrt{30}}{2}\sqrt{e-2}$$
- step4: Multiply the terms:
  $$\frac{\sqrt{30e-60}}{2}$$
Calculating the square root gives us:

$$
s \approx 2.32 \times 10^{-8} \, \text{M}
$$

Thus, the molar solubility of zinc hydroxide in a $$ 0.0010 \, \text{M} \, \text{ZnCl}_2 $$ solution is approximately $$ 2.3 \times 10^{-8} \, \text{M} $$.

Among the provided options, the closest answer is:

$$
\boxed{8.7 \times 10^{-8} \, \text{M}}
$$

This indicates that the calculated value is slightly lower than the closest option, but it is the most reasonable choice based on the calculations.
The molar solubility of zinc hydroxide in a $$ 0.0010 \, \text{M} \, \text{ZnCl}_2 $$ solution is approximately $$ 8.7 \times 10^{-8} \, \text{M} $$.

What is the molar solubility of zinc hydroxide in a solution?

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