# The molar solubility of pbBr2 at 25 °C is 1.0×10−2mol/l. Calculate ksp.

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 °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  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 $mol/L$.

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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