Cosider the dissociation of silver carbonate: Ag2CO3 (s) —-M, and [CO32-]=2.15-10-42.15-10-4 M Which of the following is true? -2Ag+ (aq) + CO32- (aq) (Ksp = 8.10-10-12 8.10-10-12) When the molarities of ions are: [Ag+] = 2.30-10-4 2.30-10-4
Please show and explain steps! Thanks!
Cosider the dissociation of silver carbonate:
Ag2CO3 (s) 2Ag+ (aq) + CO32 (
aq) (Ksp = 8.10-10-13
When the molarities of ions are:
[Ag+] = 2.30-10-4M, and [CO32] = 2.15-10-4M
Which of the following is true?
The solution does not have enough ions to form precipitate.
The solution is saturated with maximum amount of ions (but no precipitation).
The solution is supersaturated and precipitation will follow.
The Correct Answer and Explanation is :
Let’s break this down step by step to determine the correct answer.
1. Understanding the Dissociation of Silver Carbonate
The dissociation equation for silver carbonate (( \text{Ag}_2\text{CO}_3 )) is as follows:
[
\text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq)
]
The equilibrium constant, ( K_{sp} ), for this reaction is given as:
[
K_{sp} = 8.0 \times 10^{-12}
]
2. Using the Given Ion Concentrations
We are given the ion concentrations:
- ( [\text{Ag}^+] = 2.30 \times 10^{-4} \, \text{M} )
- ( [\text{CO}_3^{2-}] = 2.15 \times 10^{-4} \, \text{M} )
3. Calculating the Ion Product (( Q_{sp} ))
The ion product (( Q_{sp} )) is calculated in the same way as ( K_{sp} ), but using the actual concentrations of the ions in the solution:
[
Q_{sp} = [\text{Ag}^+]^2 \times [\text{CO}_3^{2-}]
]
Substituting the values:
[
Q_{sp} = (2.30 \times 10^{-4})^2 \times (2.15 \times 10^{-4})
]
[
Q_{sp} = (5.29 \times 10^{-8}) \times (2.15 \times 10^{-4})
]
[
Q_{sp} = 1.14 \times 10^{-11}
]
4. Comparing ( Q_{sp} ) to ( K_{sp} )
Now, compare the ion product ( Q_{sp} ) to the solubility product constant ( K_{sp} ):
- ( K_{sp} = 8.0 \times 10^{-12} )
- ( Q_{sp} = 1.14 \times 10^{-11} )
Since ( Q_{sp} ) is greater than ( K_{sp} ), this means the solution is supersaturated. A supersaturated solution has more ions than it can hold in equilibrium, which leads to precipitation.
5. Conclusion
Since ( Q_{sp} > K_{sp} ), the solution is supersaturated, and precipitation will follow. Therefore, the correct answer is:
The solution is supersaturated, and precipitation will follow.
This occurs because the concentration of ions exceeds the equilibrium concentration defined by ( K_{sp} ), leading to the formation of a precipitate as the system tries to reach equilibrium.