This question aims to find the **molar solubility constant** $ K_{sp} $ when **the molar solubility** of $PbBr _ 2$ is $ 1.0 \times 10 ^ { -2 } mol/L $ at a room temperature of **25 Â°C.**

The** molar solubility constant** is a constant represented by $k_{sp}$ which tells the amount of salt **dissolved** in a **saturated solution**. For example, if **NaCl** in the ratio of **1:1** Â is dissolved in water, it means $ Na ^ { +} $ and $ Cl ^ {-1}$ ions are present in water. We usually determine the solubility of any** salt per liter** of the saturated solution. The unit to represent the molar solubility constant is $ mol/L $.

## Expert Answer

Molar solubility of $ PbBr _ 2 $ is given by $ 1.0 \times 10 ^ { -2 } mol/L $. We will find the molar solubility constant of $ pbBr _ 2 $.

The value of $ k_{sp}$ having the general formula is determined by $ AX _ 2 $:

\[ K _ sp = 4 s ^ 3 \]

Here, **s** is the **molar solubility** of the compound.

By substituting the value of molar solubility of $ PbBr _ 2 $ in the above formula, we get:

\[ K _ sp = 4 \times ( 1.0 \times 10 ^ { -2 } ) ^ 3 \]

\[ K _ sp = 4 . 0 \times 10 ^ { – 6 } \]

## Numerical Solution

**The molar solubility constant of $ PbBr _ 2 $ is $ 4 . 0 \times 10 ^ { -6 } $.**

## Example

If the amount of $ AgIO _ 3 $ dissolved per liter of the solution is **0.0490 g** then find the molar solubility constant of $ AgIO _ 3 $.

First, we have to find the moles of $ AgIO _ 3 $ by the formula:

\[ n _ {AgIO_3 } = \frac { m } { M } \]

**M** is the **molar mass** of $ AgIO _ 3 $

**m** is the **given mass** of $ AgIO _ 3 $

The molar mass of $ AgIO _ 3 $ is **282.77 g/mol.**

Putting the values in the above formula:

\[ n _ {AgIO_3 } = \frac { 0.0490 } { 282.77 g/mol } \]

\[ n _ {AgIO_3 } = 1 . 73 \times 10 ^{ -4 } \]

Hence, the molar solubility of $ AgIO _ 3 $ is $ 1 . 73 \times 10 ^{ -4 } $

The value of $ k_{sp}$ having the general formula is determined by $ AX _ 2 $:

\[ K _ sp = 4 s ^ 2 \]

By substituting the value of molar solubility of $ AgIO _ 3 $ in the above formula, we get:

\[ K _ sp = 1 . 73 \times ( 1.0 \times 10 ^ { -4 } ) ^ 2 \]

\[ K _ sp = 3 . 0 \times 10 ^ { – 8 } \]

**The molar solubility constant of $ AgIO _ 3 $ is $ 3 . 0 \times 10 ^ { – 8 } $.**

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