what is [Ag^+] in a solution formed by mixing 25.0 ml of 0.10 M AgNO3 with 50.0 ml of 1.50 M Na3PO4. Ksp for Ag3PO4=1.0x10^-21
millimols AgNO3 = mL x M = 25*0.1 = 2.5
millimols Na3PO4 = 50 x 1.5 = 75
......3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3
I......2.5......75..........0........0
C.....-2.5.....-2.5/3......2.5/3....2.5
E.......0.......74.17.......0.8333..2.5
For solubility product with the PO4^3- as the common ion, you have
.............Ag3PO4 --> 3Ag^+ + PO4^3-
I............solid.......0.......0
C............solid.......3x......x
E............solid.......3x......x
Then Ksp = (Ag^+)^3(PO4^3-)
(Ag^+) = solive for this
(PO4^3-) = M = mmols/mL = 74.17/75 = ?
To find the concentration of Ag+ ions ([Ag+]) in the solution, we need to determine the limiting reactant, which is the reactant that will be completely used up first. Once we know the limiting reactant, we can calculate the concentration of Ag+ ions produced from that reactant.
The balanced chemical equation for the reaction between AgNO3 and Na3PO4 is:
3 AgNO3 + Na3PO4 → Ag3PO4 + 3 NaNO3
From the equation, we can see that 3 moles of AgNO3 react with 1 mole of Na3PO4 to produce 1 mole of Ag3PO4.
Step 1: Calculate the moles of AgNO3 and Na3PO4:
Moles of AgNO3 = Volume (in liters) × Concentration
= 0.025 L × 0.10 M
= 0.0025 moles AgNO3
Moles of Na3PO4 = Volume (in liters) × Concentration
= 0.050 L × 1.50 M
= 0.075 moles Na3PO4
Step 2: Determine the limiting reactant:
To find the limiting reactant, we compare the mole ratio between AgNO3 and Na3PO4 with the actual moles of each present in the solution.
From the balanced equation, the mole ratio of AgNO3 to Na3PO4 is 3:1. This means that 3 moles of AgNO3 are required to react with 1 mole of Na3PO4.
The actual mole ratio in the solution is:
(Actual moles of AgNO3) / (Actual moles of Na3PO4)
= 0.0025 moles AgNO3 / 0.075 moles Na3PO4
= 0.0333 (approx.)
Since the actual mole ratio is less than the required ratio of 3:1, AgNO3 is the limiting reactant.
Step 3: Calculate the moles and concentration of Ag+ ions:
From the balanced equation, we know that 3 moles of AgNO3 produce 1 mole of Ag3PO4.
Since AgNO3 is the limiting reactant, all of its moles will be used to form Ag3PO4. Therefore, the mole of Ag3PO4 formed is also 0.0025 moles.
From the stoichiometry of the reaction, we can see that 1 mole of Ag3PO4 produces 3 moles of Ag+ ions.
Therefore, the moles of Ag+ ions = (0.0025 moles Ag3PO4) × (3 moles Ag+ / 1 mole Ag3PO4) = 0.0075 moles Ag+
Finally, to find the concentration of Ag+ ions in the solution, we divide the moles of Ag+ ions by the total volume of the solution (in liters):
Concentration of Ag+ ions ([Ag+]) = Moles of Ag+ ions / Total volume
= 0.0075 moles / (0.025 L + 0.050 L)
= 0.0075 moles / 0.075 L
= 0.10 M
Therefore, the concentration of Ag+ ions ([Ag+]) in the solution is 0.10 M.