Question

Find the pH of a 30.0 ml butanoic acid ( HA) solution ( 0.600 M ) when you add

the following volume of 0.600 M NaOH ( 5 points ). Ka of HA = 1.5 x 10^-5

(a) 0.0 ml

(b) 10.0 ml

(c) 15.0 ml

(d) 30.0 ml

(e) 50.0 ml

Answer 1

millimoles of HA = 30 x 0.6 = 18

pKa = -log Ka = -log (1.5 x 10^-5)

pKa = 4.82

a) 0.0 ml

pH = 1/2 (pKa - log C)

    = 1/2 (4.82 - log 0.6)

pH = 2.52

b) 10.0 mL

millimoles of base = 10 x 0.6 = 6

HA + NaOH ----------------> A- +   H2O

18         6                                  0         0

12        0                                    6

pH = pKa + log [A- / HA]

    = 4.82 + log (6 / 12)

   = 4.52

pH = 4.52

c) 15.0 mL

this is half - equivalence point : here

pH = pKa

pH = 4.82

d) 30.0 mL

this is equivalence point : so here only salt remains.

salt concentration = 18 / (30 + 30) = 0.3 M

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

    = 7 + 1/2 (4.82 + log 0.3)

   = 9.15

pH = 9.15

e) 50 .0 mL

millimoles of base = 50 x 0.6 = 30

[NaOH] = 30 - 18 / (50 + 30) = 0.15 M

pOH = -log 0.15 = 0.82

pH = 13.18