Question

Using a 0.20 M phosphate buffer with a pH of 7.6, you add 0.72 mL of 0.46 M HCl to 55 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Answer 1

initially

pH = pKa + log(HPO4-2 / H2PO4-)

7.6 = 7.20 + log(HPO4-2 / H2PO4-)

(HPO4-2 / H2PO4-) = 10^(7.60-7.20) = 2.5118

(HPO4-2) = 2.5118*(H2PO4-)

(HPO4-2 + H2PO4-) = 0.2

2.5118*(H2PO4-)+( H2PO4-) = 0.2

H2PO4- = 0.2 / (2.5118+1) = 0.0569

(HPO4-2) = 2.5118*(H2PO4-) = 2.5118*0.0569 = 0.1429

now..

mmol of H2PO4- = MV = 0.0569 * 55 = 3.1295

mmol of HPO4-2 = MV = 0.1429* 55 = 7.8595

after adding

mmol of HCl = MV = 0.46*0.72= 0.3312

mmol of H2PO4- = 3.1295 + 0.3312 = 3.4607

mmol of HPO4-2 = 7.8595 - 0.3312 = 7.5283

pH = 7.21 + log(7.5283/3.4607)

pH = 7.55