Silver carbonate, Ag2CO3, has a solubility product constant of 8.46 x 10-12 at 25 °C.
(a) A sample of solid silver carbonate is added to pure water. Calculate the molar solubility (mol/L) and gram solubility (g Ag2CO3/100 mL H20) of silver carbonate in water and determine [Agt) and (CO3).
(b) Suppose 100.0 mL of 0.100 M AgNO3(aq) and 60.0 mL of 0.0600 M K2CO3(aq) are mixed. Will a precipitate form? Show your reasoning.
The Correct Answer and Explanation is :
(a) Molar Solubility of Silver Carbonate in Water
The solubility product constant (Ksp) of silver carbonate, ( \text{Ag}_2\text{CO}_3 ), is given as:
[
K_{sp} = 8.46 \times 10^{-12}
]
The dissociation of silver carbonate in water is as follows:
[
\text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq)
]
Let the molar solubility of ( \text{Ag}_2\text{CO}_3 ) be denoted by ( s ). From the dissociation equation, we can express the concentrations of the ions in solution:
- ([ \text{Ag}^+ ] = 2s)
- ([ \text{CO}_3^{2-} ] = s)
Now, using the expression for the solubility product:
[
K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}]
]
Substitute the concentrations:
[
K_{sp} = (2s)^2 \cdot s = 4s^3
]
Now solve for ( s ):
[
8.46 \times 10^{-12} = 4s^3
]
[
s^3 = \frac{8.46 \times 10^{-12}}{4} = 2.115 \times 10^{-12}
]
[
s = \sqrt[3]{2.115 \times 10^{-12}} = 1.28 \times 10^{-4} \, \text{mol/L}
]
So, the molar solubility of silver carbonate in water is ( 1.28 \times 10^{-4} \, \text{mol/L} ).
Gram Solubility of Silver Carbonate in Water
To convert the molar solubility to gram solubility, we use the molar mass of ( \text{Ag}_2\text{CO}_3 ):
[
\text{Molar mass of } \text{Ag}_2\text{CO}_3 = 2(107.87) + 12.01 + 3(16.00) = 275.87 \, \text{g/mol}
]
Now calculate the gram solubility:
[
\text{Gram solubility} = 1.28 \times 10^{-4} \, \text{mol/L} \times 275.87 \, \text{g/mol} = 0.0353 \, \text{g/L}
]
Convert this to g per 100 mL:
[
\text{Gram solubility} = 0.0353 \, \text{g/L} \times 0.1 = 0.00353 \, \text{g/100 mL}
]
Concentrations of Ions in Solution
From the molar solubility, the concentrations of the ions are:
[
[\text{Ag}^+] = 2s = 2 \times 1.28 \times 10^{-4} = 2.56 \times 10^{-4} \, \text{mol/L}
]
[
[\text{CO}_3^{2-}] = s = 1.28 \times 10^{-4} \, \text{mol/L}
]
(b) Will a Precipitate Form?
We are mixing 100.0 mL of 0.100 M AgNO₃ with 60.0 mL of 0.0600 M K₂CO₃. We first calculate the concentrations of ( \text{Ag}^+ ) and ( \text{CO}_3^{2-} ) after mixing.
- Moles of ( \text{Ag}^+ ):
[
\text{moles of Ag}^+ = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.0100 \, \text{mol}
]
Concentration of ( \text{Ag}^+ ) after mixing (total volume = 100.0 mL + 60.0 mL = 160.0 mL = 0.160 L):
[
[\text{Ag}^+] = \frac{0.0100 \, \text{mol}}{0.160 \, \text{L}} = 0.0625 \, \text{M}
]
- Moles of ( \text{CO}_3^{2-} ):
[
\text{moles of CO}_3^{2-} = 0.0600 \, \text{M} \times 0.0600 \, \text{L} = 0.00360 \, \text{mol}
]
Concentration of ( \text{CO}_3^{2-} ) after mixing:
[
[\text{CO}_3^{2-}] = \frac{0.00360 \, \text{mol}}{0.160 \, \text{L}} = 0.0225 \, \text{M}
]
Now, calculate the reaction quotient ( Q_{sp} ) and compare it to ( K_{sp} ) to determine if a precipitate will form:
[
Q_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] = (0.0625)^2 \times 0.0225 = 2.44 \times 10^{-4}
]
Since ( K_{sp} ) of ( \text{Ag}2\text{CO}_3 ) is ( 8.46 \times 10^{-12} ), and ( Q{sp} ) is much greater than ( K_{sp} ), a precipitate of ( \text{Ag}_2\text{CO}_3 ) will form.
Conclusion
- (a) The molar solubility of ( \text{Ag}_2\text{CO}_3 ) is ( 1.28 \times 10^{-4} \, \text{mol/L} ), and the gram solubility is ( 0.00353 \, \text{g/100 mL} ). The concentrations of ( \text{Ag}^+ ) and ( \text{CO}_3^{2-} ) are ( 2.56 \times 10^{-4} \, \text{mol/L} ) and ( 1.28 \times 10^{-4} \, \text{mol/L} ), respectively.
- (b) A precipitate of ( \text{Ag}2\text{CO}_3 ) will form because the reaction quotient ( Q{sp} ) is much greater than the solubility product constant ( K_{sp} ).