# Chemistry

You have at your lab bench the following chemicals: NaH2PO4(s), Na3PO4(s), NH4Cl(s), CH3COONa (s), 1.00 M NaOH, 1.00 M HCl, and deionized water. You also have standard glassware available. Describe how you would make 1.0 L of the buffer with a pH of 7.50 using only the materials listed above and deionized water so that the concentration of the acid component in the buffer is 0.100 M. Give the amounts of the pure materials in grams if they are solid or mL if they are liquid.
Acid pKa
H3PO4 2.15
H2PO4- 7.20
HPO42- 12.3
NH4+ 9.24
CH3COOH 4.75
Correct Answer: 36 g NaH2PO4, 0.20 L NaOH, water to get 1.00 L total.

I know that I have to use H2PO4- because it has the closest pH to our desired pH. In the answer key, my prof gave us I don't understand why we add NaOH and how do I find the amount of NaOH????? Please HELP!!

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1. What must base be for acid to be 0.1M?
7.5 = 7.2 + log b/a
7.5= 7.2 + log (b/0.1)
base = 0.2M

Then H2PO4^- + OH^- ==> HPO4^2- + H2O
I...........?...............0.................0..........
C.........-x.............-x....................x
E.........0.1............0.....................x

But you know x (the base) must be 0.2M so you work backwards on the ICE chart and C is -0.2M; therefore, I (initial) must be 0.1 + 0.2 = 0.3 M. So you weigh out 0.3 mol NaH2PO4 and that in 1.0 L is 0.3M. You add 0.2 mol NaOH (it is 1.0 M so that is 0.2L) and tht should give you a pH of 7.5 BUT you always check it.
..................H2PO4^- + OH^- ==> HPO4^2- + H2O
I..................0.3................0................0
C................-0.2..........-0.2....................0.2
E..................0.1..............0...................0.2

pH = 7.2 + log (b/a)
pH = 7.2 + log (0.2/0.1)
pH = 7.2 + log 2 = 7.2 + 0.3 = 7.5
Voila!

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