# Homework Help: PChem: Calculating pH

1. Dec 7, 2013

### Jormungandr

1. The problem statement, all variables and given/known data
8.9985 g of Na2HPO4 and 4.0211 g of NaH2PO4 * 7 H2O are added to 1 L of 0.20 M KCl in water. KCl does not participate in any way but to maintain ionic strength, at 0.20 M.

H3PO4 (aq) + H2O → H2PO4- (aq) + H3O+, Ka1 = 7.11 x 10-3

Ka2 = 6.32 x 10-8
Ka3 = 7.1 x 10-13

(a) Give the mass balance for phosphate, and overall charge balance, ignoring KCl.

(b) Calculate the pH.

(c) After adding 4 g NaOH, what is the new pH?

3. The attempt at a solution

So for problems like these, the setup is pretty much everything, so I wanted to make sure that I do at least the setup correct, as the rest of the problem falls out rather easily.

I started out by calculating the molarity of Na2HPO4 and NaH2PO4:

Na2HPO4: 141.96 g/mol, so 0.0634 mol. In 1 L solution, means 0.0634 M.
NaH2PO4 * 7 H2O: 246.12 g/mol, so 0.0163 mol. In 1 L solution, means 0.0163 M.

Therefore, mass balance:
0.0634 M + 0.0163 M = [H2PO4-] + [HPO42-] + [PO43-]

Charge balance:
[Na+] + [H3O+] = [H2PO4-] + 2 [HPO42-] + 3 [PO43-]

But [Na+] = 0.0163 M, so charge balance:
0.0163 M + [H3O+] = [H2PO4-] + 2 [HPO42-] + 3 [PO43-]

To find pH, I simply have to find a numerical expression for H3O+, and then do pH = –log[H3O+].

For part c, adding 4 g of NaOH changes the concentration of [Na+] ions in solution, which would change the expression of charge balance that I have, so I would just have to rework it with a different number for that value.

I haven't gone through with the calculations yet, but I just want to know if I have the gist and setup of the problem correct. Thanks in advance!

EDIT: I should add that we are given a table with hydrated radii of the ions, hence the statement about ionic strength, but this doesn't change too much other than the fact that I need to use the Debye–Hückel equation to find the activity coefficients.

2. Dec 7, 2013

### Staff: Mentor

Last edited: Dec 7, 2013
3. Dec 7, 2013

### epenguin

I hope you realise from the K's that you are going to have from that mixture H2PO4- and HPO42- as nearly all the phosphate present with [PO43-] negligible. If however you are only mentioning this last for the purpose of saying you will ignore it you should also mention H3PO4 for the purpose of then explicitly ignoring that too..

Last edited: Dec 8, 2013
4. Dec 7, 2013

### Jormungandr

Yeah, I know there's going to be very little [PO43-], but our prof wants us to calculate it as if there will be. And yeah, he does want it done the long way. It's a terrible equation, but wolfram alpha helped me get the final answer, which, if I did everything right, should be about 7.351.

5. Dec 8, 2013

### epenguin

So you are asked about a relatively refined aspect, but before that it seems to me you have made a major oversight or inconsistency, maybe only a typo? [Na+] in your charge balance equation should not be 0.0163, you seem to have forgotten the part coming form the Na2HPO4.

Last edited: Dec 8, 2013
6. Dec 8, 2013

### Staff: Mentor

Sorry for overlooking mistakes, luckily epenguin was there to correct me.

Difficult week

7. Dec 8, 2013

### Jormungandr

Yes, you're correct of course. I noticed that later on but forgot to edit it in. Also, I believe my mass balance neglected [H3PO4]. But the pH of 7.3 was after correcting for these. Oddly enough, I got around 10 after adding NaOH, however my friend got 12; not particularly sure why.

Also, does [OH] need to show up in the charge balance statement after adding NaOH?

Last edited: Dec 8, 2013
8. Dec 8, 2013

### Staff: Mentor

OH- should be present in the charge balance all the time.

9. Dec 8, 2013

### Jormungandr

Because of the autoprotolysis of water? So we'd use Kw = 1 x 10–14 to get OH- in terms of H3O+, correct? But this doesn't change the pH by an appreciable amount does it?

One thing I was struggling with was how to factor the 0.1 M contribution of the OH- into the charge balance. Does the right side of the equation change by much or is it just the left side that increases by 0.1 M because of the increased amount of Na+?

10. Dec 8, 2013

### Staff: Mentor

Yes, yes & yes.

You don't know the OH- contribution before you calculate the equilibrium, but it doesn't matter - yes, Na+ does the trick.

11. Dec 8, 2013

### epenguin

Assuming we are still talking about the first of the questions, I mentioned the [H3PO4] before, and as a matter of fact it should be of the same order of magnitude as [PO43-] . I can say this because the pK's you quote (well translating your K 1 and 3 into pK's) are about equally distant from your pH of 7.5. The two errors are in opposite directions so partly cancel each other.

You don't say how you calculated the pH taking into account just your [PO43-] . If this was an 'absolute' calculation, with some polynomial equation I guess, I would at least need to be fresher before taking it on.

Just ignoring the minor forms and considering you have in solution forms H2PO4- and HPO42- only, then the pH is an easy standard calculation. Call that first approximation - all needed for almost all practical purposes. Then assuming the pH is that, you could calculate how much the other two forms are (they will be of the order of 10-4 of the concentrations of the main forms). Then subtract the amount of the minor forms from the amount of the main forms, keeping the ratio of the latter constant. Or better maybe, now you know what [PO43-] is, write out the equation for the sum of all the forms,
0.0797 = [PO43-]*(a cubic in [H+]).
and with the computer or hand calculator solve for [H+]. Maybe you did something like this.
Actually you can calculate that by hand almost as well. Since you already have a fairly good approximation to [H+], call it [H+]1 Then let [H+] = ([H+]1 + δ) and maybe you can see how (binomial expansion) you get a good second approximation equation linear in δ.

Previously we had a calculation good enough for nearly all practical purposes, now we have one good enough for absolutely all practical purposes anyone will ever need. But you could if so inclined go through another cycle of computation starting with the second approximation values. I'd say that in order to see what these calculations are telling you it would be better to do even the first one for whch you gave pH = 7.351 to full precision of the calculator.

Note that these calculations are dealing only with the error caused by assuming the phosphate having only two forms; the error caused by ignoring [H+] and [OH-] in balance equations I think is going to be orders of magnitude smaller even than this.

Last edited: Dec 9, 2013
12. Dec 10, 2013

### epenguin

This speck of dust that had surrounded my brain blew away and I saw altogether how to do the calculation. For the fastest way it is still good to start with an approximate pH so I went to calculate that first.

There appears to be a large error in your value of 7.3 i.e. not far above the pK of 7.2.
pH = pK when molarities of acid and base forms are equal. Assuming your calculation of molarities correct, the ratio of the basic to the more acid phosphate ion is nearly 4. It did not sound right that with such a disproportion you were only 0.1 pH units above the pK. (It is useful to have at least one point on a mental map of logs between 0 and 10, and I remember that 0.5 is about log(3), so you should find yourself more than 0.5 above 7.2). The approximate pH from simple calculation e.g. Henderson-Hasselbalch taking into account only two phosphate forms I get to be 7.78

If there are signs you are still here I might go into the complete calculation, though I've already said a good bit of it.

Last edited: Dec 10, 2013
13. Dec 11, 2013

### Staff: Mentor

Strange. 0.2 M KCl can't be used to maintain ionic strength at 0.2, as solution has ionic strength around 0.2 by itself. With KCl ionic strength goes up to 0.4.

Sodium dihydrogen phosphate is not available as heptahydrate, I have found only mono and dihydrate mentioned in my handbook. There is a heptahydrate listed, but of disodium hydrogenphosphate.