What is the silver ion concentration in a solution prepared by mixing 433 mL of 0.356 M silver nitrate with 467 mL of 0.574 M sodium phosphate?
The Ksp of silver phosphate is 2.8 × 10^-18
This is a limiting reagent, solubility product, and common ion effect problem all rolled into one.
3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3.
mols AgNO3 = M x L = ? (apparox 0.15 but you should confirm the actual number.
mols Na3PO4 = M x L = ? (about 0.27)
Convert mols AgNO3 to mols of the product using the coefficients in the balanced equation.
0.15 x (1/3) = about 0.05
Convert mols Na3PO4 to mols Ag3PO4 = about 0.27
In limiting reagent problems the SMALLER value ALWAYS wins; therefore, we have approximately 0.05 mols Ag3PO4 formed.
How much Na3PO4 is used. That is
0.15 mol AgNO3 x 1/3 = about 0.05 mols Na3PO4.
How much Na3PO4 remains unreacted. That is 0.27-0.05 = about .22 mols. The concentration of the Na3PO4 is 0.22/(total volume in L) = ?M
So we have a saturated solution of Ag3PO4 with an excess of ?M Na3PO4.
........Ag3PO4 ==> 3Ag^+ + PO4^3-
..........x.........3x......x
Ksp = (Ag^+)^3(PO4^3-)
For Ag substitute 3x
For PO4^3- substitute x (for PO4 from Ag3PO4) + ?M from excess PO4^3- from Na3PO4. Solve for 3x = Ag^+
Thank you so much!!!
I am doing something wrong in my calculations.
For ?M I have 0.24
My equation is:
2.8*10^(-18) = (3x)^3 * (x+0.24)
I figured out what I was doing wrong! I have the right answer!
Yay!
Good for you.
Hey i m not getting how you did the last part how did you solve for this equation
2.8*10^(-18) = (3x)^3 * (x+0.24)
what was your answer
To find the silver ion concentration in the solution, we first need to determine which compound will precipitate out as silver phosphate. According to the given Ksp value, silver phosphate is not very soluble and will likely precipitate.
The balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium phosphate (Na3PO4) is:
3AgNO3 + Na3PO4 → Ag3PO4 + 3NaNO3
From the equation, we can see that one mole of silver nitrate reacts with one mole of sodium phosphate to form one mole of silver phosphate. Therefore, the number of moles of silver phosphate formed is the same as the number of moles of silver nitrate in the original solution.
To find the number of moles of silver nitrate, we can use the formula:
moles = concentration × volume (in liters)
For silver nitrate:
moles of AgNO3 = 0.356 M × (433 mL / 1000 mL/L) = 0.153896 moles
Since the stoichiometry of the reaction is 1:1, this means that 0.153896 moles of silver phosphate will precipitate. To find the concentration of silver ions in the solution, we divide the moles of silver ions by the total volume of the solution.
Total volume of solution = 433 mL + 467 mL = 900 mL = 0.900 L
Concentration of silver ions = moles of Ag+ / volume of solution
Concentration of silver ions = 0.153896 moles / 0.900 L = 0.1710 M
Therefore, the silver ion concentration in the solution is 0.1710 M.