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 find the ratio of moles of `Mg(OH)₂` and `Al(OH)₃` in a 1L saturated solution, we will use their solubility products (`K_sp`) and the dissociation equations for both compounds.

### Step-by-Step Solution:

1. **Write the dissociation equations:**
- For Magnesium Hydroxide:
\[ Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-} \]
- For Aluminum Hydroxide:
\[ Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} \]

2. **Define solubility:**
- Let the solubility of `Mg(OH)₂` be \( x \) moles/L.
- Then, the concentration of `Mg^{2+}` ions will be \( x \) and the concentration of `OH^{-}` ions from `Mg(OH)₂` will be \( 2x \).
- Let the solubility of `Al(OH)₃` be \( y \) moles/L.
- The concentration of `Al^{3+}` ions will be \( y \) and the concentration of `OH^{-}` ions from `Al(OH)₃` will be \( 3y \).

3. **Total concentration of `OH^{-}` ions:**
- The total concentration of `OH^{-}` ions in the solution will be:
\[ [OH^{-}] = 2x + 3y \]

4. **Write the expressions for \( K_{sp} \):**
- For `Mg(OH)₂`:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 = x(2x)^2 = 4x^3 \]
Given \( K_{sp} = 4 \times 10^{-12} \):
\[ 4x^3 = 4 \times 10^{-12} \implies x^3 = 10^{-12} \implies x = 10^{-4} \]

- For `Al(OH)₃`:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 = y(3y)^3 = 27y^4 \]
Given \( K_{sp} = 1 \times 10^{-33} \):
\[ 27y^4 = 1 \times 10^{-33} \implies y^4 = \frac{1 \times 10^{-33}}{27} \implies y^4 = \frac{10^{-33}}{27} \implies y = \left(\frac{10^{-33}}{27}\right)^{1/4} \]
\[ y = \frac{10^{-33/4}}{3^{3/4}} = \frac{10^{-8.25}}{3^{0.75}} \approx 10^{-8.25} \times 0.33 \approx 3.3 \times 10^{-9} \]

5. **Calculate the ratio of moles:**
- Now we have \( x = 10^{-4} \) and \( y \approx 3.3 \times 10^{-9} \).
- The ratio of moles of `Mg(OH)₂` to `Al(OH)₃` is:
\[ \frac{x}{y} = \frac{10^{-4}}{3.3 \times 10^{-9}} \approx \frac{10^{-4}}{3.3 \times 10^{-9}} \approx 3.03 \times 10^{5} \]

6. **Adjust the ratio as per the question:**
- The question asks to report the answer by multiplying by \( 10^{-18} \):
\[ 3.03 \times 10^{5} \times 10^{-18} = 3.03 \times 10^{-13} \]

### Final Ratio:
- The final ratio of moles of `Mg(OH)₂` to `Al(OH)₃` is approximately \( 3.03 \times 10^{-13} \).