This question aims to develop the understanding of **solubility product $ k_{ sp } $** which is related to the **solubility reactions and proportions**.

To solve this question, we may use a **four-step process**.

Step (1) – **Molar mass** estimation of the subject compound employing its **chemical formula**.

Step (2) – **Mass (in grams)** estimation of the subject compound that is **dissolved per unit liter** of the solution.

Step (3) – Number of moles estimation of the **subject compound **that is **dissolved per unit liter** of the solution.

Step (4) – Finally the **solubility product** estimation of the subject solution.

**Let’s consider the following solubility equation**:

\[ A_{(s)} \longleftrightarrow a \ A_{(a)} \ + \ b \ B_{(a)} \]

Where the **ions A and B** are the ionic decompositions of C. Factors **a and b are the proportions** involved in the reaction. The **solubility product** can be estimated through the use of following **equation**:

\[ K_{ sp } \ = \ [ A ]^a \ \times \ [ B ]^b \]

## Expert Answer

**Step (1) – Molar mass estimation of Aluminum Hydroxide $ Al ( OH )_3 $:**

\[ \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 3 \bigg ( 1 \ + \ 16 \bigg ) \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 3 \bigg ( 17 \bigg ) \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 51 \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 78 \ g/mole \]

**Step (2) – Mass (in grams) estimation of Aluminum Hydroxide $ Al ( OH )_3 $ dissolved per unit liter or 1000 milliliter solution:**

Since its not given, lets assume it as $ x $.

**Step (3) – Number of moles estimation of Aluminum Hydroxide $ Al ( OH )_3 $ dissolved per unit liter or 1000 milliliter solution:**

\[ \text{ Moles dissolved in 1 L solution } = \ \dfrac{ \text{ Mass dissolved in 1 L solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow \text{ Moles dissolved in 1 L solution } = \ \dfrac{ x }{ 78 } \ moles \]

**Step (4) – Solubility product estimation.**

The solubility equation of given reaction can be written as follows:

\[ Al ( OH )_3 (s) \longleftrightarrow \ Al^{ +3 } ( aq ) \ + \ 3 \ OH^{ -1 } ( aq ) \]

This means that:

\[ [ Al ( OH )_3 ] \ = \ [ Al^{ +3 } ] \ = \ 3 [ OH^{ -1 } ] \ = \ \dfrac{ x }{ 78 } \ mole \]

\[ \Rightarrow [ OH^{ -1 } ] \ = \ \dfrac{ x }{ 26 } \ mole \]

So:

\[ K_{ sp } \ = \ [ Al^{ +3 } ]^1 \ \times \ [ OH^{ -1 } ]^3 \]

\[ \Rightarrow K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

## Numerical Result

\[ K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

Where x is the grams dissolved per unit liter of solution.

## Example

For the **same scenario** given above, calculate the $ K_{ sp } $ if **100 g is dissolved in a 1000 mL solution**.

**Calculating the number moles of copper chloride $ Cu Cl $ dissolved in 1 L = 1000 mL solution:**

\[ x \ = \ \dfrac{ \text{ Mass in 1000 mL solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow x \ = \ \dfrac{ 100 }{ 78 \ g/mole } \]

\[ \Rightarrow x \ = \ 1.28 \ mole/L \]

**Recall the final expression:**

\[ K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

**Substituting values:**

\[ K_{ sp } \ = \ \dfrac{ 1.28 }{ 78 } \ \times \ \bigg ( \dfrac{ 1.28 }{ 26 } \bigg )^3 \]

\[ K_{ sp } \ = \ 0.01652 \]