how would you prepare 250 mL solution containing the following: 0.30 M mannitol, 0.025 M phosphate buffer pH 7.5? You are given solid mannitol, kh2po4, and k2hpo4. (pKa=6.86)

Answer 1

I assume this means you want a phosphate buffer of pH = 7.5 containing 0.3M mannitol. So the mannitol part is to weigh out 0.3 mol mannitol/L or you will want 0.3/4 mols for the 250 mL. Then grams mannitol = mols mannitol x molar mass mannitol. Place that in a 250 mL volumetric flask.

For the phosphate buffer you solve two equation simultaneously.
Here is equation 1. The Henderson-Hasselbalch equation is
7.5 = pKa2 + log base/acid
Solve for base/acid ratio, then
base = approx 2.3 so
base = 2.3*acid. I like to work in millimols so that would be mmols base = mmols acid*2.3

Equation 2 =
acid + base = 0.025 x 250
Solve for acid and base millimols, then from
grams = mols x molar mass, calculate g KH2PO4 and g K2HPO4, weigh, place in the 250 mL volumetric flask and make to the mark. Mix thoroughly and stopper.