plzzz teaxh me this!!

1M sol of H3PO4 = 3H^+ + 1PO4^- = 3N
but then its treated with CaCL2
1Ca^2+ + 2 Cl^- = 2N
3/2= 1.5N
..... something like that 😉

hopelessdoc said:

what is the normality of a 1M sol of H3PO4 if titrated with CaCL2?

ans: 1.5 N

thanx...

Click to expand...

is that all the information they give you?

yes...this is all they had for the ques...ofcourse do had other answer choices if that's what u asking, "lifeisgood" ??...but I just wrote the correct ans choice here...

Chga said:

1M sol of H3PO4 = 3H^+ + 1PO4^- = 3N
but then its treated with CaCL2
1Ca^2+ + 2 Cl^- = 2N
3/2= 1.5N
..... something like that 😉

Click to expand...

thanx....but plzz clarify all this...
1) how the heck u figure out Normality? I know whats the general def is but form this equation why we choose the moles of H^+ as the Normality value?

2) why the Normality of CaCl^2 is 2 not 1?
😕 ...guys I think if I don't get all this basic right ...no chance for me to get anywhere...so depressed...

This question makes no sense at all, it seems like half of a question...

hopelessdoc said:

yes...this is all they had for the ques...ofcourse do had other answer choices if that's what u asking, "lifeisgood" ??...but I just wrote the correct ans choice here...

Click to expand...

The normality of an acid is defined as the number of of equivalents of H+ in one liter of solution. An equivalent is defined as the weight of an acid that produces 1 mole of H+ ions...in this problem equivalents=moles. Therefore, a 1M solution of H3PO4 is 3N wrt H. When you titrate w/ CaCl2 the PO43- will ppt with the Ca2+ (CaPO4 is insoluble)...and the volume of the solution will change giving you a larger volume, since the number of equivalents of H+ (HCl strong acid) doesn't change but the number of liters does you will get a smaller N. I can't figure out how they got 1.5 N though, doesn't make sense to me. What were the other answer choices??? Seems like you would need more info like the conc of CaCl2...

Hope this didn't confuse you more...and if anyone really knows how to do this problem I would love to know.

armorshell said:

This question makes no sense at all, it seems like half of a question...

Click to expand...

Exactly what I was thinking...driving me nuts 😕

The N of the H3PO4 is going to be 3 times the M.

CaCl2 Ca2+ + 2 Cl-
1 M CaCl2 = 2 N CaCl2

Ca chloride has two moles of chloride ions for every mole of CaCl2. Because of that, the multiplying factor for calcium chloride is two.

3/2 = 1.5 👍

That is of course assuming a 1M solution of CaCl2.

KOM said:

The N of the H3PO4 is going to be 3 times the M.

CaCl2 Ca2+ + 2 Cl-
1 M CaCl2 = 2 N CaCl2

Ca chloride has two moles of chloride ions for every mole of CaCl2. Because of that, the multiplying factor for calcium chloride is two.

3/2 = 1.5 👍

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I still don't buy it...why divide the normalities?

Okay, I figured out what this question wants.

It's asking for the normality with respect to Ca in CaCl2, to fully titrate H3PO4, which would be 1.5 normal

armorshell said:

Okay, I figured out what this question wants.

It's asking for the normality with respect to Ca in CaCl2, to fully titrate H3PO4, which would be 1.5 normal

Click to expand...

If it's with respect to Ca, than why do we still use 2N of CaCl2?? I thought its 2N of 2Cl- ions!! if it's an acid-base rxn...and only 1N with respect to Ca+2 ions...

lifeisgood said:

The normality of an acid is defined as the number of of equivalents of H+ in one liter of solution. An equivalent is defined as the weight of an acid that produces 1 mole of H+ ions...in this problem equivalents=moles. Therefore, a 1M solution of H3PO4 is 3N wrt H. When you titrate w/ CaCl2 the PO43- will ppt with the Ca2+ (CaPO4 is insoluble)...and the volume of the solution will change giving you a larger volume, since the number of equivalents of H+ (HCl strong acid) doesn't change but the number of liters does you will get a smaller N. I can't figure out how they got 1.5 N though, doesn't make sense to me. What were the other answer choices??? Seems like you would need more info like the conc of CaCl2...

Hope this didn't confuse you more...and if anyone really knows how to do this problem I would love to know.

Click to expand...

I am confused with the division part? the ans choices are: 0.5N, 1.0 N, 1.5N and 2.0 N...

hopelessdoc said:

If it's with respect to Ca, than why do we still use 2N of CaCl2?? I thought its 2N of 2Cl- ions!! if it's an acid-base rxn...and only 1N with respect to Ca+2 ions...

Click to expand...

You didn't indicate any normality, nor with what element the normality is in relation too.

I was trying to reverse engineer the solution from the answer, with just the question you posted noone could get the answer. Where's it from?

This is all they had in the ques...it is from the kaplan's one of the online quiz...I think "properties of solution" quiz section...

Okay, I posed this question to an old professor of mine who damn near genius and he thinks the questioni s ******ed... My e-mail first then his (names have been changed to protect the innocent)

Hey Dr. X,

I have a chemistry question for you that was posed to me by someone else. I cannot reason through it.

They gave me the question:

"What is the normality of a 1M solution of H3PO4 after titration with CaCl2?"

That is all the information given and he said the answer was 1.5 N. I have been trying to figure out why without any luck. I figure it must be a ppt titration b/c CaPO4 is insoluble. Since HCl is a strong acid, the number of equivalent H+ doesn't change but the volume does and I know that the normality has to decrease (from 3 N to ???) just because the volume increases. But I guess what is throwing me off is that if we assume the CaCl2 to be 1 M, you would need 3 parts CaCl2 to every 2 parts H3PO4 for complete ppt. The numbers are confusing me and it seems like the question does not have enough info.

The only way I can think to get the final normality equal to 1.5 would be to titrate one liter of 1 M H3PO4 with an equal volume of 1.5 M CaCl2. Final products would be solid CaPO4 and 1.5 M HCl and the normality with respect to H+ would be 1.5 N.

Please help!

Thanks,

lifeisgood




Lifeisgood,
I will assume that the question was supposed to be: "What is the
normality of a 1M solution of H3PO4 to be used for titration with
CaCl2?"
Normality is defined as the number of "equivalents" of solute per
liter of solution, and the number of "equivalents" will depend on the
reaction for which the solute is used. In neutralization reactions, an
equivalent of an acid is the amount in moles that supplies one mole of
H+; for redox reactions,it's the number of moles of solute that
transfers one mole of electrons; for precipitation and complex
formation reactions, it's the number of moles of solute that supplies
or reacts with (or is stoichiometrically equivalent to) one mole of
univalent cation in the precipitate or complex formed. In the case of
the reaction: 3CaCl2(aq) + 2H3PO4(aq) -> (Ca)3(PO4)2(s)+ 6HCl(aq), 2
moles of H3PO4 would react with 3 moles of divalent cation (equal to 6
moles of univalent cation), or 1 mole of H3PO4 would react with 3
moles of univalent cation. Therefore, the normality of a 1 M H3PO4
solution to be used for this reaction would be 3 N, not 1.5 N.
You're correct in thinking that the solution after titration could be
1.5 N (with respect to its subsequent use for neutralization) if it
were diluted 1:2 in the precipitation titration, but if this is what's
meant, the question is ambiguous, trickily-worded, and ia a rather
tortuous example of a normality problem.
Dr. X

And Armorshell, I will have to agree with you when you said the solution of CaCl2 would be 1.5 N wrt to Ca. That is the ONLY logical explanation.

thanx "lifeisgood" for all ur efforts and ofcourse others too!!...but still it's confusing for me...does this kinda stuff show up on dats? any idea...

hopelessdoc said:

thanx "lifeisgood" for all ur efforts and ofcourse others too!!...but still it's confusing for me...does this kinda stuff show up on dats? any idea...

Click to expand...

I have not taken the DAT yet but I have no doubt that a normality question could and would be on it. This question is just downright ambiguous. I would try to find some other practice questions on it and get a hold of a chemistry textbook or use google to search for an explanation you can digest.

It really is a simple concept when explained in the correct way. Good luck!

hopelessdoc said:

what is the normality of a 1M sol of H3PO4 if titrated with CaCL2?

ans: 1.5 N

thanx...

Click to expand...

IMO, normality of one substance depends on a particular reaction, and can be understood that the normality is the amount/mole equivelent of that substance to react with 1 mole of another reactant. In this reaction:
3CaCl2(aq) + 2H3PO4(aq) -> (Ca)3(PO4)2(s)+ 6HCl(aq)
3 moles of CaCl2 reacts with 2 moles of H3PO4. So, 1.5 moles of CaCl2 reacts with 1 mole of H3PO4. In another word, 1 mole of H3PO4 is equivelent to 1.5 moles of CaCl2. That's why the normality of H3PO4 with respect to CaCl2 is 1.5N.
Hope this will help.🙂