Question

Determine the pH of 50.0 mL of a 0.100 M solution of propanoic acid after the following additions. 0.00 mL of 0.100 M NaOH, 30.00 mL of 0.100 M NaOH, 50.00 mL of 0.100 M NaOH, 60.00 mL of 0.100 M NaOH

Answer 1

millimoles of acid = 50 x 0.1 = 5

pKa of acid = 4.89

1) 0.00 mL of 0.100 M NaOH

pH = 1/2 [pKa - logC]

pH = 1/2 [4.89 -log 0.1]

pH = 2.94

2) 30.00 mL of 0.100 M NaOH

millimoles of NaoH = 0.1 x 30 = 3

HA   +   NaOH ------------------> NaA + H2O

5           3                                  0            0

2          0                                   3             3

pH = pKa + log [NaA/ HA]

pH = 4.89 + log (3 / 2)

pH = 5.06

3) 50.00 mL of 0.100 M NaOH

it is equivalence point . here only salt remains both acid and base consumed

salt concetration = 5 / ( 50 + 50) = 0.05 M

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [4.89 + log 0.05]

pH = 8.79

4) 60.00 mL of 0.100 M NaOH

[OH-] = (6 - 5) / 110

          = 9.09 x 10^-3 M

pOH = -log [OH-]

pOH = -log (9.09 x 10^-3)

pOH = 2.04

pH + pOH = 14

pH = 11.96