Question

As a technician in a large pharmaceutical research firm, you need to produce 100. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.91. The p K a of H 2 P O 4 ? 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.)

Answer 1

The pH of a buffer solution can be determined by the Henderson-Hasselbalch equation Acid: KH2PO, Salt KHPO K2HPO4 4 KHPO4 pHpka + log pKa = 7.21 H2PO4 + H2O KH PO 2 pH 6.91 Each term is replaced by the values KHPO 6.91 7.21 +log KH PO K.HPO4 6.91 -7.21log KH PO 2 KHPO, 0.3 log КНРО Log is removed KHPO4 10-0.3 КНРО KHPO4 0.50- KHAPO For each mole of acid 0.50 mole of s salt needed to prepare the solution at the indicated pH

30 mL of 1M acid was chosen Acid: 30 mL sal: 30 mL x 0.50 = 15 mL (KH2PO4) al 30mL (acid 15 ml (sal) total 45 mL of the solution will be obtained at pH 6.91 Water is added 55 mL of water and 100 mL of buffer is obtained at pH But 100 ml is requested ! 6.91