|Jul2-12, 12:41 AM||#1|
Find pH of buffer
1. The problem statement, all variables and given/known data
A buffer is made by dissolving 13.0g of sodium dihydrogen phosphate, NaH2PO4, and 15.0g of disodium hydrogen phosphate, Na2HPO4, in a litre of solution. What is the pH of the buffer?
2. Relevant equations
pH = -log [H+]
3. The attempt at a solution
First, I'm not sure if this is right, but I assumed that NaH2PO4 dissassociates into Na+ and then H2PO4-, and then H+, H+, PO4 3- ions. Second I assumed that it dissassociates into Na+, Na+, H+ and PO4 3- ions.
M(Na2HPO4) = 141.96
3) M(NaH2PO4) = 119.98
n=.10835 * 2 = 0.2167 (since hydrogen ion concentration doubles)
4) 0.105664+0.2167=0.322 mol H+
5) pH = -log(.322) = 0.492
This is completely off. Please help. I realize I have a completely wrong approach, but I don't know what else can be done for this problem.
|Jul2-12, 03:14 AM||#2|
While HPO42- and H2PO4- do dissociate slightly, you can safely ignore it, and assume their concentrations can be calculated from known number of moles and volume.
Have you heard about Henderson-Hasselbalch equation? This is just a rearranged acid dissociation constant used in such cases.
|Jul2-12, 09:58 AM||#3|
The Henderson-Hasselbalch equation is
pH = pKa + log([A-]/[HA])
However, I don't know the pKa because no constant is given.
Gah. I still don't know what to do.
|Jul3-12, 11:08 PM||#4|
Find pH of buffer
You'll have to find the constant for NaH2PO4 in some table of values, or look online.
|Jul5-12, 05:18 AM||#5|
And to get the relevant molarities you will have to know molecular weights. Were they not given? Or the formulae including water of crystallisation, because these usually come as crystals with stoichiometric water such as Na2 HPO4.7H2O and NaH2PO4.H2O (but also sometimes sold anhydrous)? If not given you will have to state your assumption, but it seems a bit ridiculous to give a multiple choice question with incomplete starting data.
Edit: on second thoughts you can eliminate several of those options qualitatively without calculation but not yet obvious to me how you can home in on just one.
|Jul5-12, 06:09 AM||#6|
I guess OP is expected to assume anhydrous salts. That's the way these questions are usually constructed (whether it makes sense, or not). This way you end with (almost) equimolar solution.
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