What is the silver ion concentration in a solution prepared by mixing 487 mL of 0.384 M silver nitrate with 379 mL of 0.583 M sodium phosphate? The Ksp of silver phosphate is 2.8 × 10-18
This is a limiting reagent problem and a Ksp problem with a common ion all rolled into one.
3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3
mol AgNO3 = M x L = about 0.19 but you need to do it more accurately.
mols Na3PO4 = M x L = about 0.22 but you confirm.
Convert mols AgNO3 to mols Ag3PO4. That's 1/3 x mol AgNO3 = about 0.06 (the coefficients in the equation give you the fraction.)
Convert mols Na3PO4 to mols Ag3PO4. That's 0.22.
You have two values for mols Ag3PO4 and both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller of the two and the reagent responsible for that small number is the limiting reagent. Thus 0.06 mols Ag3PO4 will be formed and AgNO3 is the limiting reagent. That makes Na3PO4 the reagent in excess.
So we now have a saturated solution of Ag3PO4 with a excess of Na3PO4. How much excess? That's 0.22-(0.06/3) = about 0.2
........Ag3PO4 --> 3Ag^+ + PO4^3-
.........solid......3x......x
(Ag^+) = 3x
(PO4^3-) = x + 0.2
Ksp Ag3PO4 = (Ag^+)(PO4^3-)
Substitute the concn into Ksp expression and solve for x, then Ag^+ = 3x.
Post your work if you get stuck.
Ok so I get everything up until the end
Ksp = [Ag+]^3[Po43-] since Ag3PO4 <----> 3Ag+ + PO43-
so then i do ((3x)^3)(x+0.2) = 2.8e-18
is this correct?... it comes out to be a very complex mathematical problem to solve for x which is why i feel it's wrong
You are ok to this point. Are you sure of the 0.2? That was an estimate on my part. If that is ok, then you can solve the cubic equation or make an assumption that x + 0.2 = 0.2 and solve the resulting equation which is easy enough to do. I would make the assumption; when you finish compare 0.2 with x+0.2 and see if the assumption that x was too small to add in is ok. I think the assumption will be ok.
got it, thanks!
To find the silver ion concentration in the solution, we need to first find out how much silver phosphate will precipitate out of the solution, and then calculate the concentration of silver ions remaining.
To do this, we can start by calculating the moles of silver nitrate and sodium phosphate used in the solution.
1. Moles of silver nitrate:
Concentration of silver nitrate = 0.384 M
Volume of silver nitrate = 487 mL = 487/1000 L
Moles of silver nitrate = Concentration × Volume
= 0.384 M × (487/1000) L
2. Moles of sodium phosphate:
Concentration of sodium phosphate = 0.583 M
Volume of sodium phosphate = 379 mL = 379/1000 L
Moles of sodium phosphate = Concentration × Volume
= 0.583 M × (379/1000) L
Next, we will determine the limiting reactant by comparing the moles of silver nitrate and sodium phosphate. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
3. Compare the moles of silver nitrate and sodium phosphate:
Let's assume the moles of silver nitrate is 'a' and moles of sodium phosphate is 'b'.
a = 0.384 M × (487/1000) L (from step 1)
b = 0.583 M × (379/1000) L (from step 2)
The ratios a:b will reveal the limiting reactant.
If a/b is less than 3/4 (or 0.75), then silver nitrate is the limiting reactant.
If a/b is more than 3/4 (or 0.75), then sodium phosphate is the limiting reactant.
Now, let's assume silver nitrate is the limiting reactant. The balanced equation for the reaction between silver nitrate and sodium phosphate is:
3 AgNO3(aq) + Na3PO4(aq) → Ag3PO4(s) + 3 NaNO3(aq)
From the balanced equation, we can see that 3 moles of silver nitrate react with 1 mole of sodium phosphate to form 1 mole of silver phosphate.
4. Moles of silver phosphate formed:
Moles of silver phosphate = a/3 (since 3 moles of silver nitrate form 1 mole of silver phosphate)
Using the moles of silver phosphate formed, we can calculate its concentration in the resulting solution.
5. Volume of resulting solution:
Total volume of solution = Volume of silver nitrate + Volume of sodium phosphate
= 487 mL + 379 mL
= (487 + 379)/1000 L
6. Concentration of silver phosphate:
Concentration of silver phosphate = Moles of silver phosphate / Volume of resulting solution
Finally, we can calculate the concentration of silver ions in the solution by considering the dissociation of silver phosphate.
7. Silver ion concentration:
Since all the silver phosphate will dissociate into silver ions (Ag+) and phosphate ions (PO4^-3), the concentration of silver ions is equal to the concentration of silver phosphate.
Therefore, the silver ion concentration in the solution is equal to the concentration of silver phosphate calculated in step 6.
Please note that this calculation assumes that silver phosphate is the only solid precipitate formed and that there are no other reactions or complexation occurring in the solution.