What is the concentration of Ag+ remaining in solution?
Calculate the following when 0.250 L of 0.249 M AgNO3 solution is mixed with 0.350 L of 0.300 M Na2CO3 solution.
Ksp for Ag2CO3 is 8.46E-12.
A) What mass of the precipitate forms?
I found out the answer: 8.59 g.
But I don't know how to solve B, I attempted it and got it wrong.
B) What is the concentration of Ag+ remaining in solution?
B.
Let x = solubility of Ag2CO3
Ag2CO3 ==> 2Ag^+ + CO3^2-
...x........2x.......x
Ksp = (Ag^+)^2(CO3^2-)
You know Ksp or can look it up. Calculate how much of the CO3^2- was used in part A and subtract from the initial amount CO3^2- in the problem. Substitute this into Ksp expression for CO3^2- and solve for x = Ag2CO3, then multiply that by 2 to find 2x = (Ag^+).
To solve part B) and find the concentration of Ag+ remaining in solution, we need to calculate the initial number of moles of Ag+ ions and determine how many moles of Ag+ ions have reacted to form Ag2CO3.
First, let's find the initial number of moles of Ag+ ions in the AgNO3 solution.
We have 0.250 L of 0.249 M AgNO3 solution.
Number of moles of Ag+ = concentration (M) x volume (L)
Number of moles of Ag+ = 0.249 M x 0.250 L
Next, let's determine the number of moles of Ag+ that react with Na2CO3 to form Ag2CO3.
From the reaction equation, we know that the molar ratio between Ag+ and Ag2CO3 is 2:1.
So, the number of moles of Ag+ ions that react = (1/2) x number of moles of Ag+ ions.
Now, let's calculate the number of moles of Ag+ ions that react:
Number of moles of Ag+ ions that react = (1/2) x number of moles of Ag+ ions
Finally, to find the concentration of Ag+ ions remaining in the solution, we need to subtract the number of moles of Ag+ ions that react from the initial number of moles of Ag+ ions.
Concentration of Ag+ remaining = (number of moles of Ag+ ions - number of moles of Ag+ ions that react) / total volume of solution
Substituting the values, we can now solve for the concentration of Ag+ remaining in solution.
To find the concentration of Ag+ remaining in solution, we need to determine the amount of Ag+ that reacts with the Na2CO3 solution, and then calculate the remaining concentration.
Here's how you can solve it:
1. Write the balanced chemical equation for the reaction between AgNO3 and Na2CO3:
AgNO3 + Na2CO3 → Ag2CO3 + 2NaNO3
2. Calculate the moles of AgNO3 and Na2CO3 using the given volumes and molarities:
Moles of AgNO3 = volume (L) × molarity (M) = 0.250 L × 0.249 M
Moles of Na2CO3 = volume (L) × molarity (M) = 0.350 L × 0.300 M
3. Determine the limiting reagent by comparing the moles of AgNO3 and Na2CO3. The reactant that produces fewer moles of Ag2CO3 is the limiting reagent.
4. Use stoichiometry to calculate the moles of Ag2CO3 formed. The mole ratio between AgNO3 and Ag2CO3 is 1:1.
5. Calculate the mass of Ag2CO3 precipitate formed by multiplying the moles of Ag2CO3 by its molar mass. This is the answer you found for part A (8.59 g).
Now, to solve part B:
6. Calculate the initial moles of Ag+ ions present in the AgNO3 solution by multiplying the moles of AgNO3 (from step 2) by the mole ratio between Ag+ and AgNO3, which is also 1:1.
7. Calculate the moles of Ag+ ions that reacted with the Na2CO3 solution by multiplying the moles of Ag2CO3 formed (from step 4) by the mole ratio between Ag+ and Ag2CO3, which is 2:1.
8. Calculate the remaining moles of Ag+ ions by subtracting the moles of Ag+ that reacted (from step 7) from the initial moles of Ag+ (from step 6).
9. Finally, calculate the concentration of Ag+ remaining in solution by dividing the moles of Ag+ remaining by the total volume of the solution.
Note: Make sure to pay attention to significant figures and unit conversions throughout the calculations.
By following these steps, you should be able to find the correct answer for part B.