The solubility of Ag2CrO4 in water at 25°C is measured to be 0.022g/L . Use this information to calculate Ksp for Ag2CrO4 . Round your answer to 2 significant digits.
The correct answer and explanation is :
To calculate the solubility product constant (Ksp) for ( \text{Ag}_2\text{CrO}_4 ), we first need to use the solubility of ( \text{Ag}_2\text{CrO}_4 ) in water and relate it to the equilibrium concentrations of the ions in solution.
Step 1: Write the dissociation equation
When silver chromate (( \text{Ag}_2\text{CrO}_4 )) dissolves in water, it dissociates into silver ions (( \text{Ag}^+ )) and chromate ions (( \text{CrO}_4^{2-} )):
[
\text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq)
]
Step 2: Determine molar solubility
The given solubility is 0.022 g/L, which represents the amount of ( \text{Ag}_2\text{CrO}_4 ) that dissolves in 1 liter of water.
To convert this to molar solubility, we need the molar mass of ( \text{Ag}_2\text{CrO}_4 ).
The molar mass of ( \text{Ag}_2\text{CrO}_4 ) is calculated as:
- ( 2 \times \text{Ag} = 2 \times 107.87 = 215.74 \, \text{g/mol} )
- ( \text{Cr} = 52.00 \, \text{g/mol} )
- ( 4 \times \text{O} = 4 \times 16.00 = 64.00 \, \text{g/mol} )
Total molar mass = ( 215.74 + 52.00 + 64.00 = 331.74 \, \text{g/mol} )
Now, we convert the solubility from grams to moles:
[
\text{Molar solubility} = \frac{0.022 \, \text{g/L}}{331.74 \, \text{g/mol}} = 6.63 \times 10^{-5} \, \text{mol/L}
]
Step 3: Express concentrations of ions
From the dissociation equation, for each mole of ( \text{Ag}_2\text{CrO}_4 ) that dissolves, we get:
- 2 moles of ( \text{Ag}^+ )
- 1 mole of ( \text{CrO}_4^{2-} )
Thus, the concentrations at equilibrium will be:
- ( [\text{Ag}^+] = 2 \times 6.63 \times 10^{-5} = 1.33 \times 10^{-4} \, \text{mol/L} )
- ( [\text{CrO}_4^{2-}] = 6.63 \times 10^{-5} \, \text{mol/L} )
Step 4: Write the expression for Ksp
The solubility product constant ( K_{sp} ) for ( \text{Ag}2\text{CrO}_4 ) is given by: [ K{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}]
]
Substitute the equilibrium concentrations:
[
K_{sp} = (1.33 \times 10^{-4})^2 \times (6.63 \times 10^{-5})
]
[
K_{sp} = 1.77 \times 10^{-8}
]
Step 5: Final Answer
So, the solubility product constant for ( \text{Ag}2\text{CrO}_4 ) is: [ K{sp} = 1.8 \times 10^{-8} \, (\text{rounded to two significant digits})
]
Explanation:
The solubility product constant (Ksp) is a measure of the extent to which a compound will dissolve in water. By using the solubility of ( \text{Ag}_2\text{CrO}_4 ), we were able to calculate the molar solubility and then determine the concentrations of the dissociated ions at equilibrium. From this, we used the Ksp expression to calculate the value of the solubility product constant.