posted by moses on .
As a technician in a large pharmaceutical research firm, you need to produce 200.mL of 1.00 M a phosphate buffer solution of pH = 7.19. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?
You need 200 mL x 1M so
base(b) + acid(a) = 0.2 mols.
That's equation 1.
7.19= 7.21 + log b/a
b/a = ? and that's equation 2
Solve those two equations to find a and b then substitute each into M = mol/L to find the individual amounts for a and b.
Post your work if you get stuck.
I went through this in a hurry and came up with about 98 mL base and 103 mL acid but these are not exact.