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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 $.

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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