Question

How do you prepare one liter of phosphate buffer 0.01 M at pH 12.0 from 85% phosphoric acid (density 1.69g/ml) and NaOH 1.0 M

Answer 1

A fairly long solution.
Start with two equations. The first is to find the ratio of base to acid.
pH = pK3 + log (base)/(acid)
Substitute and solve for b/a.
Equation 2 is b + a = 0.01.
Solve those two equations simultaneously for a (acid) and b (base). Since those are M concns and you want a L those will be mols in a L. I estimate mols acid = about 0.007 and mols base = about 0.003

How do you get 0.007 mols H3PO4. That will be 0.007 (remember to do these calculations more accurately than my estimates) x 98 g/mol = approx 0.7 g.
How to get approx 0.7 grams H3PO4.
mL x density x %acid = 0.7 and solve for mL H3PO4.
For the base, you want approx 0.003 mols. M = mols/L. YOu know M and mols, solve for L and convert to mL if needed.

There is another way to do this and the problem may be asking for that solution.
Calculate the a and b as above, then calculate mL H3PO4. Place in a beaker, add water, stir, insert pH meter electrodes into the solution and add 1.0 M NaOH until the pH reads 12.0. Dilute to 1 L.

Post your work if you get stuck.