# college chem

You are instructed to create 400. mL of a 0.40 M phosphate buffer with a pH of 6.9. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.)

H3PO4(s) + H2O(l) equilibrium reaction arrow H3O+(aq) + H2PO4−(aq)
Ka1 = 6.9 ✕ 10−3

H2PO4−(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + HPO42−(aq)
Ka2 = 6.2 ✕ 10−8

HPO42−(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + PO43−(aq)
Ka3 = 4.8 ✕ 10−13

What is the molarity needed for the acid component of the buffer?

What is the molarity needed for the base component of the buffer?

How many moles of acid are needed for the buffer?

How many moles of base are needed for the buffer?

How many grams of acid are needed for the buffer?

How many grams of base are needed for the buffer?

I really can't figure this out. I tried the ph=pka +log(a-/ha) but I'm really lost and if you could guide be that would be wondering. I'm really stressing out here.

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1. Here is the hint you should need to work the problem. You have two equations and you solve them simultaneously (which I think is the easier way) but there are other ways to do it.
pH = pKa + log (base)/(acid)
Take the Ka of k1, k2, and k3. The closes one to 6.9 is the one you want and that is k2.
6.9 = pK2 + log b/a
b/a = ? . That's the first equation.
The second one is
a + b = 0.4M
Solve for a and b (in molarity). That will answer the first two. Convert to mols - M x L = ? and that gives the next two answers. The mols = grams/molar mass gives you the last two answers.

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