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?
9 answers
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.
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.
1 answer