Question: you have .420 Litters of a solution containing Ag3PO4 (s) in equilibrium with its ions. write an algebraic equation which when solved will geive the molar concentration of Ag^+1 after the addition of .200 moles of Na3Po4(s) to the .420 liters of solution. in your equation, x must equal to the equilibrium molar concentration of Ag^+1 in the solution after the addition of Na3PO4 (s). ksp for Ag3PO4= 1.3E-20.

So the equation would be Ag3PO4 ==> 3Ag + PO4. wouldn't the equation look like 1.3E-20= (3x)^3 (.4762+x)? Is this set up correctly or am i doing somthing wrong?

Your equation looks great; however, the problem states that x is to equal to (Ag^+). You have it set up so that x = (Ag3PO4) = (PO4^-3) and (Ag^+) = 3x

I'm a little confused on what that means. does that mean that .4762 would be + 3x instead of X?

No. The phosphate ion is 0.4762 (if your prof is a freak about significant digits perhaps this should be 0.48 since only 2 places are shown for the Ksp in the problem. The problem states that x = (Ag^+) [in fact it is repeated in the problem] and what you have is (Ag^+) = 3x.

So the equation would really look like 1.3E-20=(3X)^3 (.4762). or am i still doing somthing wrong?

There is nothing wrong with the equation except that when you solve for x you get what? What did you let x stand for? It looks to me as if you let x stand for the solubility of Ag3PO4 and Ag^+ is then 3x. So when you solve your equation, you will get x THEN YOU MUST MULTIPLY IT BY 3 TO GET 3x. And the problem specifically states that x is to be the Ag^+ not 3x = Ag^+. So your answer of x is not what the problem asks. Right?

So if i'm solving for 3x how do i set up the problem to solve for x

Try something here.

Work the problem as (3x)^3(0.476) = 1.3E-20, solve for x, then multiply by 3 and that will be Ag^+.
NOW, try letting Ag^+ = x (instead of 3x).
How will the equation look then? Will it be (x)^3(0.476) = 1.3E-20.
Solve that and see if x the second time is the same as 3x the first time.

I will try to post those sites under a new post. Look at the top for them.

The sites won't post.

To solve this problem, we need to set up an equilibrium expression and then solve for the molar concentration of Ag^+1. The equation you wrote is almost correct, but there is a small mistake. Let's break down the steps to help you set it up correctly:

Step 1: Write the balanced chemical equation:
Ag3PO4(s) ⇌ 3 Ag^+(aq) + PO4^3-(aq)

Step 2: Write the expression for the equilibrium constant:
Ksp = [Ag^+]^3[PO4^3-]

Given that Ksp for Ag3PO4 is 1.3 × 10^-20, we can write the expression as:
1.3 × 10^-20 = [Ag^+]^3[PO4^3-]

Step 3: Determine the stoichiometry of the reaction:
From the balanced equation, we know that for every 1 mole of Ag3PO4 that dissolves, 3 moles of Ag^+ are produced. Thus, the concentration of Ag^+ will be 3 times higher than the concentration of Ag3PO4.

Therefore, we can express the concentration of Ag^+ in terms of the concentration of Ag3PO4 as:
[Ag^+] = 3x, where x is the molar concentration of Ag3PO4.

Step 4: Determine the new molar concentration of Ag^+ after the addition of Na3PO4:
When Na3PO4(s) is added, it will react with Ag^+ to form Ag3PO4(s) precipitate. The amount of Ag^+ that will react can be determined by stoichiometry.

Since 3 moles of Ag^+ ions react with 1 mole of Na3PO4, the molar concentration of Ag^+ will decrease by 3 times the molar concentration of Na3PO4 added.

Since 0.200 moles of Na3PO4 are added to 0.420 liters of solution, the molar concentration of Na3PO4 is:
[Na3PO4] = 0.200 moles / 0.420 liters = 0.4762 M

Therefore, the decrease in the molar concentration of Ag^+ is:
-3 × 0.4762 M = -1.4286 M

So, the new molar concentration of Ag^+ after the addition of Na3PO4 will be:
[Ag^+] (new) = [Ag^+] (equilibrium) + (-1.4286 M)

Step 5: Substitute the expression for [Ag^+] in the equilibrium expression and solve for x:
1.3 × 10^-20 = (3x - 1.4286)^3

Solve this equation for x to find the equilibrium molar concentration of Ag^+ after the addition of Na3PO4.

I hope this explanation helps you set up the correct algebraic equation and solve the problem!