Question

How many grams of NaOH(Mr 40) must be added to 100 ml of 0.1M H3PO4 to create phosphate buffer, pH 7.0?

I know I'm using HH-equation. H3PO4, pka1=2.12, pka2=7.21, pka3=12.32.

Heres what I got:

pH=Pka+ log HPO4/H3PO4-HPO4
7.0=7.21 + log HPO4/.001-HPO4

is this right dr.Bob...I'm solving for HPO4, right?
Isn't a phosphate buffer @ pH about 7.2 composed of NaH2PO4 and Na2HPO4?
So you must add enough NaOH to completely neutralize the first hydrogen; therefore, you add 100mL x 0.1M mmoles NaOH to START. Then calculate the base(HPO4--/H2PO4-) you need to ADD to the NaOH initially used to arrive at the desired pH. .
100 ml * 0.1 M= 10 mmole

7.0 = 7.21 + log HPO4^-2/ 10- HPO4^-2

HPO4^-2 = 1.27M

how do i get to grams

Right, Dr.Bob?

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