Question

As a technician in a large pharmaceutical research firm, you need to produce 350. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.91. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)

Express your answer to three significant digits with the appropriate units. Please show all steps.

Answer 1

Solution :-

Lets first calculate the ratio of the conjugate acid to base using the Henderson equation

pH= pka + log [base/acid]

6.91 = 7.21 + log [base/acid]

6.91-7.21 = log [base/acid]

-0.30 = log [base/acid]

Antling [-0.30] = [base/acid]

0.50 = [base/acid]

0.50 = x/(1-x)

Therefore

x= 0.333

so the concentration of the base is 0.333 M

so the concentration of the acid = 1-x = 1-0.333 M = 0.667 M

now lets calculate the moles of the acid and conjugate base that are present in 350 ml = 0.350 L solution

H2PO4- moles = 0.667 mol per L * 0.350 L = 0.23345 mol

HPO4^2- mol = 0.333 mol per L * 0.350 L = 0.11655 mol

Now using the moles and the molarities of the stock solutions lets calculate the volume of the stock solution needed

Volume of H2PO4- = moles / molarity

                                 = 0.23345 mol / 1.0 mol per L

                                = 233 ml

233 ml * 1 L / 1000 ml = 0.233 L

So the volume of the KH2PO4 needed is 0.233 L