This question aims to find the **solubility product $ k_{ sp } $** involved in the **solubility reactions and proportions**.

This is a **four-step process**. First, we find the **molar mass of the given compound** using its chemical formula. Second, we find the **mass of of given compound **dissolved in 1 L solution. Third, we find the number moles of **given compound** dissolved in 1 L solution. Fourth, we find the **solubility product of the solution**.

**Given a reaction**:

\[ A_{(s)} \longleftrightarrow d \ B_{(a)} \ + \ e \ C_{(a)} \]

Where **B and C are the ions** formed as a result of dissolving A while **d and e are the proportions**. The **solubility product** can be calculated by using the following **formula**:

\[ K_{ sp } \ = \ [ B ]^d \ \times \ [ C ]^e \]

## Expert Answer

**Step (1) – Calculating the molar mass of copper chloride $ Cu Cl $:**

\[ \text{Molar mass of CuCl } = \ \text{Molar mass of copper } + \text{ Molar mass of chlorine } \]

\[ \Rightarrow \text{Molar mass of CuCl } = \ 63.546 \ + \ 35.453 \]

\[ \Rightarrow \text{Molar mass of CuCl } \ = \ 98.999 \ \approx \ 99 \ g/mole \]

**Step (2) – Calculating the mass of of copper chloride $ Cu Cl $ dissolved in 1 L = 1000 mL solution:**

\[ \text{ 100 mL of copper chloride } = \ 3.91 \ mg \]

\[ \Rightarrow \text{ 1 mL of copper chloride } = \ \dfrac{ 3.91 }{ 100 } \ mg \]

\[ \Rightarrow \text{ 1000 mL of copper chloride } = \ 1000 \times \dfrac{ 3.91 }{ 100 } \ mg \ = \ 39.1 \ mg \]

\[ \Rightarrow \text{ 1000 mL of copper chloride } \ = \ 39.1 \ mg \ = \ 0.0391 \ g \]

**Step (3) – Calculating the number moles of copper chloride $ Cu Cl $ dissolved in 1 L = 1000 mL solution:**

\[ \text{ Number of Moles in 1000 mL solution } = \ \dfrac{ \text{ Mass in 1000 mL solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow \text{ Number of Moles in 1000 mL solution } = \ \dfrac{ 0.0391 }{ 99 \ g/mole } \]

\[ \Rightarrow \text{ Number of Moles in 1000 mL solution } = \ 0.000395 \ mole \]

**Step (4) – Calculating the solubility product constant $ K_{ sp } $.**

The solubility reaction can be written as:

\[ CuCl \longleftrightarrow Cu^+ \ + \ Cl^- \]

This means that:

\[ [ CuCl ] \ = \ [ Cu^+ ] \ = \ [ Cl^- ] \ = \ 0.000395 \ mole \]

So:

\[ K_{ sp } \ = \ [ Cu^+ ]^1 \ \times \ [ Cl^- ]^1 \]

\[ \Rightarrow K_{ sp } \ = \ 0.000395 \ \times \ 0.000395 \]

\[ \Rightarrow K_{ sp } \ = \ 1.56 \times 10^{ -7 } \]

## Numerical Result

\[ K_{ sp } \ = \ 1.56 \times 10^{ -7 } \]

## Example

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

**Step (1)** – We already have the **molar mass** of **copper chloride **$ Cu Cl $.

**Step (2)** – The **mass** of of **copper chloride **$ Cu Cl $ dissolved in 1 L = 1000 mL solution is given.

**Step (3)** – Calculating the **number of moles** of **copper chloride **$ Cu Cl $ dissolved in 1 L = 1000 mL solution:

\[ \text{ Number of Moles in 1000 mL solution } = \ \dfrac{ \text{ Mass in 1000 mL solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow \text{ Number of Moles in 1000 mL solution } = \ \dfrac{ 100 \ g }{ 99 \ g/mole } \]

\[ \Rightarrow \text{ Number of Moles in 1000 mL solution } = \ 1.01 \ mole \]

**Step (4)** – Calculating the **solubility product constant** $ K_{ sp } $:

\[ [ CuCl ] \ = \ [ Cu^+ ] \ = \ [ Cl^- ] \ = \ 1.01 \ mole \]

So:

\[ K_{ sp } \ = \ [ Cu^+ ]^1 \ \times\ [ Cl^- ]^1 \ = \ 1.01 \ \times\ 1.01 \ = \ 1.0201 \]