Question

construct by calculation a titration curve for the titration of 25.00 ml of .250 M CH3CH2COOH (Pka= 4.89) with .330 M NaOH. Specifically what is a) the colume of titrant required to reach the equivalence point ; and what is the pH b) initially, c) after the addition of 2.00 m; of .330 M NaOH, d) at the half neutralization point, e) aftr the addition of 12.00ml of .330 M NaOH, f)at the equivalence point, and g) after the addition of 20.00 ml of .330 M NaOH?

Answer 1

millimoles of acid = 25 x 0.250 = 6.25

pKa = 4.89

a) the colume of titrant required to reach the equivalence point :

At equivalence point :

millimoles of acid = millimoles of base

6.25 = 0.330 x V

V = 18.94 mL

volume of titrant = 18.94 mL

b) initially pH :

pH = 1/2 (pKa - log C)

     = 1/2 (4.89 - log 0.250)

pH = 2.75

c) after the addition of 2.00 m; of .330 M NaOH :

millimoles of base = 2 x 0.330 = 0.66

CH3CH2COOH   + NaOH   ------------------> CH3CH2COONa + H2O

   6.25                    0.66                                   0

5.59                      0                                       0.66

pH = pKa + log [salt / acid]

     = 4.89 + log [0.66 / 5.59]

pH = 3.96

d) at the half neutralization point

At half neutralization point :

pH = pKa

pH = 4.89

e) aftr the addition of 12.00ml of .330 M NaOH

millimoles of base = 12 x 0.330 = 3.96

CH3CH2COOH   + NaOH   ------------------> CH3CH2COONa + H2O

   6.25                    3.96                                  0

2.29                      0                                       3.96

pH = pKa + log [salt / acid]

     = 4.89 + log [3.96 / 2.29]

pH = 5.13