Calculate the maximum solubility of silver carbonate, Ag2CO3 in g/L when in the presence of 0.055 M Na2CO3. The solubility product of Ag2CO3 is 7.2×10 12 and Ag2CO3 has a molar mass of 275.75 g/mol. Express your answer to the correct number of significant figures, in scientific notation and include the unit with your answer.
The Correct Answer and Explanation is :
To calculate the maximum solubility of silver carbonate (Ag2CO3) in the presence of Na2CO3, we will need to take into account the solubility product constant (Ksp) and the common ion effect. Here’s a step-by-step solution:
Step 1: Understanding the Dissolution Process
The dissociation of silver carbonate (Ag2CO3) in water can be represented by the following equation:
[
\text{Ag}2\text{CO}_3 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq) ] The equilibrium expression for the solubility product (Ksp) of Ag2CO3 is: [ K{\text{sp}} = [\text{Ag}^+]^2 [\text{CO}3^{2-}] ] Given that ( K{\text{sp}} = 7.2 \times 10^{-12} ), we will use this value to calculate the solubility of Ag2CO3.
Step 2: Considering the Common Ion Effect
In this case, sodium carbonate (Na2CO3) is added to the solution. Na2CO3 dissociates completely in water to give CO3²⁻ ions:
[
\text{Na}_2\text{CO}_3 (aq) \rightarrow 2\text{Na}^+ (aq) + \text{CO}_3^{2-} (aq)
]
The concentration of CO3²⁻ ions from Na2CO3 is 0.055 M, which will shift the equilibrium of the Ag2CO3 dissolution process. The concentration of CO3²⁻ ions due to Na2CO3 is already 0.055 M, and we will denote the concentration of Ag⁺ from Ag2CO3 as ( s ). The equilibrium concentration of CO3²⁻ from Ag2CO3 dissociation will be ( s ).
Step 3: Writing the Equilibrium Expression
The solubility product expression for Ag2CO3 in the presence of Na2CO3 becomes:
[
K_{\text{sp}} = (2s)^2 \cdot (0.055 + s)
]
Since the concentration of CO3²⁻ from Na2CO3 is much higher than the amount that will dissociate from Ag2CO3 (because 0.055 M is much larger than ( s )), we can approximate:
[
K_{\text{sp}} = (2s)^2 \cdot 0.055
]
This simplifies to:
[
K_{\text{sp}} = 4s^2 \cdot 0.055
]
Step 4: Solving for ( s )
Now, substitute the value for Ksp and solve for ( s ):
[
7.2 \times 10^{-12} = 4s^2 \cdot 0.055
]
[
s^2 = \frac{7.2 \times 10^{-12}}{4 \times 0.055}
]
[
s^2 = \frac{7.2 \times 10^{-12}}{0.22}
]
[
s^2 = 3.27 \times 10^{-11}
]
[
s = \sqrt{3.27 \times 10^{-11}} \approx 5.72 \times 10^{-6} \, \text{M}
]
Step 5: Converting to Solubility in g/L
To find the solubility in grams per liter, we need to use the molar mass of Ag2CO3 (275.75 g/mol):
[
\text{Solubility} = s \times \frac{\text{Molar mass of Ag}_2\text{CO}_3}{2}
]
Since 2 moles of Ag⁺ are produced per mole of Ag2CO3, the solubility in terms of Ag2CO3 is:
[
\text{Solubility} = 5.72 \times 10^{-6} \, \text{M} \times \frac{275.75 \, \text{g/mol}}{2}
]
[
\text{Solubility} = 5.72 \times 10^{-6} \times 137.875 \, \text{g/L}
]
[
\text{Solubility} = 7.89 \times 10^{-4} \, \text{g/L}
]
Final Answer:
The maximum solubility of Ag2CO3 in the presence of 0.055 M Na2CO3 is:
[
\boxed{7.89 \times 10^{-4} \, \text{g/L}}
]
This is the solubility of Ag2CO3 under the given conditions, expressed in scientific notation with the correct number of significant figures.