Two equations which you solve simultaneously. KH2PO4 is the acid(A); K2HPO4 is the base(B)
pH = pKa + log B/A
So equation 1 is
7.20 = 7.21 + log B/A
Solve for B/A = ?
Equation 2 is
A + B = 1 M. (The problem tells you the buffer is to be 1 M. I like to work in millimols and you want 450 mL of 1 M; therefore,
A + B = 450 mmols.
Solve these two equations
(1) B/A = ?
(2) A + B = 450
for A and B. A is approx 230 mmols and B is approx 220 mmols.
You will need better answers for each since these numbers are estimates (but close estimates)
Now you know M = mmols/mL. You know M for A and mmolsl for A, solve for mL A.
And you know M for B and mmols for B, solve for mL B.
My estimates are aprox 150 mL for B and 230 mL for A. Again these are close estimates. Post your work if you get stuck.
As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M a phosphate buffer solution of pH = 7.20. 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?
1 answer
1 answer