What is the ratio of moles of `Mg(OH)_(2)` and `Al(OH)_(3)`, present in 1L saturated solution of `Mg(OH)_(2)` and `Al(OH)_(3) K_(sp)` of `Mg(OH)_(2)=4xx10^(-12)` and `K_(sp)` of `Al(OH)_(3)=1xx10^(-33)`.[Report answer by multiplying `10^(-18)]`

Explanation

To solve the problem, we need to find the ratio of moles of `Mg(OH)₂` and `Al(OH)₃` in a saturated solution, given their solubility product constants (`Ksp`).

### Step-by-Step Solution:

1. **Identify the Dissociation Reactions**:
- For `Mg(OH)₂`:
\[ Mg(OH)₂ \rightleftharpoons Mg^{2+} + 2OH^{-} \]
If we let the solubility of `Mg(OH)₂` be \( x \) moles/L, then:
- \( [Mg^{2+}] = x \)
- \( [OH^{-}] = 2x \)

- For `Al(OH)₃`:
\[ Al(OH)₃ \rightleftharpoons Al^{3+} + 3OH^{-} \]
If we let the solubility of `Al(OH)₃` be \( y \) moles/L, then:
- \( [Al^{3+}] = y \)
- \( [OH^{-}] = 3y \)

2. **Write the Ksp Expressions**:
- For `Mg(OH)₂`:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 = x(2x)^2 = 4x^3 \]
Given \( K_{sp} = 4 \times 10^{-12} \):
\[ 4x^3 = 4 \times 10^{-12} \implies x^3 = 10^{-12} \implies x = 10^{-4} \text{ moles/L} \]

- For `Al(OH)₃`:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 = y(3y)^3 = 27y^4 \]
Given \( K_{sp} = 1 \times 10^{-33} \):
\[ 27y^4 = 10^{-33} \implies y^4 = \frac{10^{-33}}{27} \implies y = \left(\frac{10^{-33}}{27}\right)^{1/4} \]

3. **Calculate y**:
- First, calculate \( \frac{10^{-33}}{27} \):
\[ \frac{10^{-33}}{27} \approx 3.70 \times 10^{-35} \]
- Now, take the fourth root:
\[ y = (3.70 \times 10^{-35})^{1/4} = 10^{-8.75} \text{ moles/L} \approx 5.62 \times 10^{-9} \text{ moles/L} \]

4. **Find the Ratio of Moles**:
- The ratio \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{10^{-4}}{5.62 \times 10^{-9}} \approx 1.78 \times 10^{5} \]

5. **Multiply by \( 10^{-18} \)**:
- Finally, multiply the ratio by \( 10^{-18} \):
\[ 1.78 \times 10^{5} \times 10^{-18} = 1.78 \times 10^{-13} \]

### Final Answer:
The ratio of moles of `Mg(OH)₂` to `Al(OH)₃`, reported by multiplying by \( 10^{-18} \), is approximately \( 1.78 \times 10^{-13} \).

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