What is the silver ion concentration in a solution prepared by mixing 495 mL 0.369 M silver nitrate with 419 mL 0.471 M sodium phosphate? The 𝐾sp of silver phosphate is 2.8×10−18.
millimoles AgNO3 = mL x M = 495 x 0.369 = about 182.6
millimoles Na3PO4 = 419 x 0.471 = about 197.3
You should check these and correct to the correct number of significant figures.
............3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3
I............182.6..........197.3.................0...............0
This is a limiting reagent problem as well as a solubility product problem.
How much Ag3PO4 is formed and how much of the reactants remain.
182.6 mmols AgNO3 x (1 mol Ag3PO4/3 mols AgNO3) = 60.9
197.3 mmols Na3PO4 x (1 mol Ag3PO4)/1 mol Na3PO4) = 197.4 so the LR is AgNO3 (all of it is used) with Na3PO4 excess remaining.
I............182.6..........197.3.................0...............0
C.........-182.6..........-60.9................+60.9 ........182.6
E..............0.............136.4..................60.9.......................
So we have a solution with 60.9 mmols of Ag3PO4 ppt (solid) with 0 mmols AgNO3 and 60.9 mmols Na3PO4.
The (Ag^+) will be determined by the common ion effect of the phosphate.
Ag3PO4 ==> 3Ag^+ + PO4^3-
Ksp = 2.8E-18 = (Ag^+)^3(PO4^3-)
Look above. (Ag^+) = solve for this.
(PO4^3-) = mmols Na3PO4 from above/ total volume in mL
Note that total volume is 495 mL + 419 mL = ? mL.
Check all of this to confirm each step.
To determine the silver ion concentration in the solution, we need to find the limiting reactant first by comparing the number of moles of silver nitrate and sodium phosphate.
1. Calculate the number of moles of silver nitrate:
Moles of AgNO3 = Volume (in liters) x Molarity
= 0.495 L x 0.369 M
= 0.182655 moles
2. Calculate the number of moles of sodium phosphate:
Moles of Na3PO4 = Volume (in liters) x Molarity
= 0.419 L x 0.471 M
= 0.197349 moles
3. Next, we need to determine the ratio between AgNO3 and Na3PO4 based on their balanced equation. The balanced equation for the reaction between silver nitrate and sodium phosphate is:
3 AgNO3 + Na3PO4 → Ag3PO4 + 3 NaNO3
4. From the balanced equation, we can see that a 3:1 molar ratio exists between AgNO3 and Na3PO4. However, since the moles of sodium phosphate (0.197349) are less than one-third of the moles of silver nitrate (0.182655), we consider Na3PO4 as the limiting reactant.
5. Convert the limiting reactant moles (0.197349) to moles of silver phosphate produced using the balanced equation (1 mole of Ag3PO4 is produced for every 1 mole of Na3PO4):
Moles of Ag3PO4 = Moles of Na3PO4 = 0.197349 moles
6. Now, using the molar mass of Ag3PO4, we can calculate the grams of Ag3PO4 formed:
Molar mass of Ag3PO4 = 108.77 g/mol
Mass of Ag3PO4 = Moles of Ag3PO4 x Molar mass of Ag3PO4
= 0.197349 moles x 108.77 g/mol
= 21.45 g
7. Finally, we can calculate the silver ion concentration in the solution using the mass and the volume of the solution:
Concentration of silver ions = Mass of Ag3PO4 / Volume of solution
= 21.45 g / (0.495 L + 0.419 L)
= 21.45 g / 0.914 L
= 23.45 g/L or 23.45 mg/mL
Therefore, the silver ion concentration in the solution is 23.45 mg/mL.
To determine the silver ion concentration in the solution, we can use the concept of stoichiometry and the solubility product constant (Ksp) of silver phosphate.
1. Start by writing the balanced equation for the reaction between silver nitrate (AgNO3) and sodium phosphate (Na3PO4), which forms silver phosphate (Ag3PO4):
3AgNO3 + Na3PO4 → Ag3PO4 + 3NaNO3
2. Based on the balanced equation, we can see that for every 3 moles of silver nitrate, we get 1 mole of silver phosphate. Therefore, the mole ratio is 1:3.
3. Calculate the number of moles of silver nitrate used:
Moles of AgNO3 = volume (in L) × molarity
= 0.495 L × 0.369 M
= 0.1827 moles AgNO3
4. Determine the number of moles of silver phosphate formed using the mole ratio:
Moles of Ag3PO4 = Moles of AgNO3 × (1 mole Ag3PO4 / 3 moles AgNO3)
= 0.1827 moles / 3
= 0.0609 moles Ag3PO4
5. Calculate the concentration of silver phosphate in the final solution:
Volume of final solution = sum of volumes of both solutions used
= 0.495 L + 0.419 L
= 0.914 L
Concentration of Ag3PO4 = Moles of Ag3PO4 / Volume of final solution
= 0.0609 moles / 0.914 L
= 0.0668 M Ag3PO4
6. Since silver phosphate is a sparingly soluble salt, it will dissociate into silver ions (Ag+) and phosphate ions (PO4^3-) according to the stoichiometry of the balanced equation.
7. The solubility product constant (Ksp) describes the equilibrium between the solid salt and its ions in a saturated solution. For silver phosphate, the Ksp value is given as 2.8 × 10^(-18).
8. Since the stoichiometry of the reaction is 1:3 between silver nitrate and silver phosphate, the concentration of silver ions (Ag+) will be three times the concentration of silver phosphate. Therefore:
Concentration of Ag+ = 3 × Concentration of Ag3PO4
= 3 × 0.0668 M
= 0.2004 M
So, the silver ion concentration in the solution is 0.2004 M.