Question

A buffer that contains 0.22 M of a base, B and 0.28 M of its conjugate acid BH+, has a pH of 9.87. What is the pH after 0.02 mol of Ba(OH)2 are added to 0.62 L of the solution?

Answer 1

B = 0.22

HB+ = 0.28

pH = 9.87

pOH = 14-9.87 = 4.13

pOH = pKb + log(HB+/B)

solve for pKB

pKb = pOH -log(HB+/B) = 4.13 - log(0.28/0.22) = 4.02526

then

after adding Ba(OH)2

0.02*2 = 0.04 mol of OH-

mol of B initially = MV = 0.62*0.22 = 0.1364 mol of B

mol of HB+ initially = MV = 0.62*0.28 = 0.1736 mol of HB+

then

mol of B after= 0.1364 +0.04 = 0.1764

mol of HB+ after=0.1736 -0.04 = 0.1336

then

pOH = pKb + log(HB+/B)

pOH = 4.02526 + log(0.1336/0.1764) = 3.9045

pH = 14-3.9045 = 10.0955