Question

Show the full calculations necessary in order to calculate the pH of the resulting solution after the following volumes of 0.30 M sodium hydroxide are added to 30.00 mL of 0.15 M acetic acid. a) 0.00 mL b) 10.25 mL c) 15.00 mL d) 22.00 mL

Answer 1

pKa of acetic acid = 4.74

a)

pH = 1/2 (pKa - log C)

     = 1/2 (4.74 - log 0.15)

pH = 2.78

b)

millimoles of acetic acid = 30 x 0.15 = 4.5

millimoles of NaOH = 0.30 x 10.25 = 3.075

CH3COOH +   NaOH   -----------> CH3COONa   + H2O

    4.5               3.075                           0                    0

1.425                 0                            3.075

pH = pKa + log [salt / acid]

     = 4.74 + log [3.075 / 1.425]

pH = 5.07

c)

millimoles of NaoH = 15 x 0.3 = 4.5 .

this is equivalence point.here salt only remains.

salt concentration = 4.5 / (30 + 15) = 0.1 M

pH = 7 +1/2 (pKa + log C)

     = 7 + 1/2 (4.74 + log 0.1)

pH = 8.87

d)

millimoles of NaOH = 22 x 0.3 = 6.6

[OH-] = 6.6 - 4.5 / 22 + 30 = 0.0404 M

pOH = 1.39

pH = 12.61