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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