A solution is prepared by mixing 45.00 mL of 0.022 M AgNO3 with 13.00 mL of 0.0014 M Na2CO3. Assume volumes are additive.
Calculate [Ag+], [CO32-], [Na+] and [NO3-] after equilibrium is established.
....2AgNO3 + Na2CO3 ==> Ag2CO3 + 2NaNO3
mmols AgNO3 = 45.00*0.022 = 0.99
mmols Na2CO3 = 13.00*0.0014 = 0.0182.
As a limiting reagent (LR) problem we could form 0.0182 mol Ag2CO3 with Na2CO3 or 0.495 mol Ag2CO3 with AgNO3. That makes Na2CO3 the LR. So we use up all of the Na2CO3 (at least all of the CO3^2-) and 0.0182*2 = 0.0364 mol AgNO3 which leaves 0.99-0.364 = 0.954 mols AgNO3 in excess.
I assume you are not to worry about Ksp of Ag2CO3(s) since the Ag^+ from AgNO3 is a common ion and will affect solubility of Ag2CO3.
Total volume is 58 mL. We are assuming all of the carbonate is gone.
(Ag^+) = 0.954/58 = ?
etc.
To calculate the concentrations of the ions after equilibrium is established, we need to determine the number of moles of each ion present in the given volumes of the solutions.
First, let's calculate the number of moles of AgNO3 and Na2CO3 in each solution:
For AgNO3:
Number of moles = volume (in L) × molarity
Number of moles of AgNO3 = 45.00 mL × (1 L / 1000 mL) × (0.022 mol/L)
Number of moles of AgNO3 = 0.00099 mol
For Na2CO3:
Number of moles = volume (in L) × molarity
Number of moles of Na2CO3 = 13.00 mL × (1 L / 1000 mL) × (0.0014 mol/L)
Number of moles of Na2CO3 = 0.0000182 mol
Now, let's determine the moles of each ion present after mixing the solutions:
Since AgNO3 dissociates into Ag+ and NO3-, and Na2CO3 dissociates into 2Na+ and CO32-, we can write the balanced equation for the reaction as:
AgNO3 + Na2CO3 -> Ag+ + 2Na+ + CO32- + NO3-
Looking at the equation, we can see that 1 mole of AgNO3 produces 1 mole of Ag+ and 1 mole of NO3-
Similarly, 1 mole of Na2CO3 produces 2 moles of Na+ and 1 mole of CO32-
Thus, the moles of each ion after mixing can be calculated as follows:
[Ag+] = moles of Ag+ / total volume of solution (in L)
[NO3-] = moles of NO3- / total volume of solution (in L)
[Na+] = moles of Na+ / total volume of solution (in L)
[CO32-] = moles of CO32- / total volume of solution (in L)
Now, let's substitute the values:
[Ag+] = 0.00099 mol / (45.00 mL + 13.00 mL) × (1 L / 1000 mL)
[Ag+] = 0.00099 mol / 0.058 L
[Ag+] ≈ 0.017 mol/L
[NO3-] = 0.00099 mol / (45.00 mL + 13.00 mL) × (1 L / 1000 mL)
[NO3-] = 0.00099 mol / 0.058 L
[NO3-] ≈ 0.017 mol/L
[Na+] = 2 × (0.0000182 mol) / (45.00 mL + 13.00 mL) × (1 L / 1000 mL)
[Na+] = 0.0000364 mol / 0.058 L
[Na+] ≈ 0.00063 mol/L
[CO32-] = 0.0000182 mol / (45.00 mL + 13.00 mL) × (1 L / 1000 mL)
[CO32-] = 0.0000182 mol / 0.058 L
[CO32-] ≈ 0.00031 mol/L
So, after equilibrium is established, the concentrations of [Ag+], [CO32-], [Na+], and [NO3-] in the solution will be approximately 0.017 M, 0.00031 M, 0.00063 M, and 0.017 M, respectively.