A 20.0 gram sample is a mixture of sodium phosphate ,sodium mono-hydrogen phosphate,sodium dihydrogen phosphate and sodium chloride. The sample is dissolved in 100.0 ml of deionized water and titrated with 1.5 M hydrochloric acid. The initial pH of the solution is pH= 12.9. The first equivalence point occurs at 16.0 ml of acid ( pH= 10.5). The second equivalence occurs after the addition of an additional 44.0 ml of acid( pH= 8.05). The third equivalence occurs after the addition of an additional 11.1 ml of acid( pH= 1.22). Calculate the mass percent of each component in the mixture.
3 answers
See my response to your earlier post of this problem.
it isn't showing your response in the earlier post!
I must have hit the wrong button when I tried to post. Here are my partial thoughts. I said something like,
Bill, we can't draw diagrams on the board and I don't know how to explain a problem like this without better aids than are available. I can try to get you started with some info and you can re-post with very specific questions if you have trouble. This really is a long and complicated problem.
The first thing you need to know is/are the equation(s).
PO4^3- is titrated first, HPO4^2- is second, and H2PO4^- is last.
PO4^3- + H^+ ==> HPO4^2-
HPO4^2- + H^+ ==> H2PO4^-
H2PO4^- + H^+ ==> H3PO4.
The first equivalence point is straight forward. mL x M = 16.0 x 1.5M = ? millimoles HCl = same mmols PO4^3-. Then you convert that to grams and you have the first component.
The problem gets more complicated from here.
You have mL for the second equivalence point BUT that is the volume to titrate the original HPO4^2- in the sample PLUS the HPO4^2- formed in getting to the first equivalence point. You will need to subtract an equivalent amount to find HPO4^2- in the original sample. I would calculate total millimoles - millimoles for the first amount = millimoles for the second amount (in the initial 20g sample) Then the third one as twice as many problems as this. The third equivalence point is the volume to titrate the original H2PO4^- + H2PO4^- formed from the first stage + H2PO4^- from the second stage.
Good luck. Post your work if you get stuck but better yet be VERY explicit about what you don't understand.
Bill, we can't draw diagrams on the board and I don't know how to explain a problem like this without better aids than are available. I can try to get you started with some info and you can re-post with very specific questions if you have trouble. This really is a long and complicated problem.
The first thing you need to know is/are the equation(s).
PO4^3- is titrated first, HPO4^2- is second, and H2PO4^- is last.
PO4^3- + H^+ ==> HPO4^2-
HPO4^2- + H^+ ==> H2PO4^-
H2PO4^- + H^+ ==> H3PO4.
The first equivalence point is straight forward. mL x M = 16.0 x 1.5M = ? millimoles HCl = same mmols PO4^3-. Then you convert that to grams and you have the first component.
The problem gets more complicated from here.
You have mL for the second equivalence point BUT that is the volume to titrate the original HPO4^2- in the sample PLUS the HPO4^2- formed in getting to the first equivalence point. You will need to subtract an equivalent amount to find HPO4^2- in the original sample. I would calculate total millimoles - millimoles for the first amount = millimoles for the second amount (in the initial 20g sample) Then the third one as twice as many problems as this. The third equivalence point is the volume to titrate the original H2PO4^- + H2PO4^- formed from the first stage + H2PO4^- from the second stage.
Good luck. Post your work if you get stuck but better yet be VERY explicit about what you don't understand.
0 answers