Question

Determine the pH during the titration of 24.3 mL of 0.310 M formic acid (Ka = 1.8×10-4) by 0.442 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 4.20 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 25.6 mL of NaOH

Answer 1

(a) Before the addition of any NaOH

First, assume the acid:

Hformic

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.31 M; then

x^2 + (1.8*10^-4)x - 0.3*(1.8*10^-4) = 0

solve for x

x =0.00738

substitute

[H+] = 0 + 0.00738= 0.00738M

pH = -log(H+) = -log(0.00738) = 2.13

pH = 2.13

(b) After the addition of 4.20 mL of NaOH

mmol of acid = MV = 0.31*24.3 = 7.533

mmol of base = MV = 0.442*4.20 = 1.8564

then

pH = pKa + log(A-/HA)

pH = 3.75 + log( (1.8564) / (7.533-1.8564))

pH = 3.264

(c) At the half-equivalence point (the titration midpoint)

in half equivalnece point

A- = HA

so

pH = pKa + log(a--/HA)

pH = pKa + log(1)

pH = pKa

pH = 3.75

(d) At the equivalence point

in equivalece point

Vbase = mmol/V = 24.3*0.31/0.442 = 17.042

Vtotal = 24.3+17.042 = 41.342 mL

[A-] = mmol/V = 7.533/41.342= 0.182

Kb = [HA][OH-]/[A-]

5.5*10^-11 = x+x/(0.182-x)

x = 3.18*10^-6

pH = 14 + log(OH) = 14 + log(3.18*10^-6) =8.50

(e) After the addition of 25.6 mL of NaOH

excess base

mmol of base = MV = 25.6*0.442 = 11.325

mmol of acid = 7.533

mmol of Oh- left = 11.325-7.533 =) 3.792

pH = 14 + log(3.792 / (24.3+25.6) = 12.88