Question

6. You have 20.00 mL of a 0.100 M aqueous solution of the weak base (CH3)3N (Kb = 7.40 x 10-5). This solution will be titrated with 0.100 M HCl.

(a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added?
(c) What is the pH of the solution after 5.00 mL of acid has been added? (d) What is the pH of the solution at the equivalence point of the titration? (e) What is the pH of the solution when 25.00 mL of acid has been added?

Answer 1

a)

The reaction between (CH3)3N and HCl is 1:1 molar

(CH3)3N + HCl ---------> (CH3)3NH⁺ + Cl^-

M1V1 = M2V2

V2 = M1V1/M2

= 0.100M × 20.00ml/0.100M

= 20.00ml

Therefore ,

Volume of acid required to reach end point = 20.00ml

b)

(CH3)3N + H2O <--------> (CH3)3NH⁺ + OH^-

K_b = [(CH3)3NH⁺] [OH^-]/[(CH3)3N] = 7.40 ×10^-5

at equilibrium

[(CH3)3N] = 0.100 - x

[(CH3)3NH⁺] = x

[OH^-] = x

x²/( 0.100 -x) = 7.40 ×10^-5

solving for x

x = 0.002684

[OH^-] = 0.002684M

pOH = -log[OH^-]

pOH = -log( 0.002684M)

pOH = 2.57

pH = 14 - pOH

pH = 14 - 2.57

pH = 11.43

c)

Initial moles of (CH3)3N = (0.100mol/1000ml) × 20.00ml = 0.0020mol

moles of HCl added = (0.100mol/1000ml) × 5.00ml = 0.0005mol

0.0005 moles of (CH3)3N react with 0.0005moles of HCl to give 0.0005moles of (CH3)3NH⁺

remaining moles of (CH3)3N = 0.0020mol - 0.0005mol = 0.0015mol

After addition of HCl

Total volume = 25.00ml

[(CH3)3N] = ( 0.0015mol/ 25.00ml) ×1000ml = 0.0600M

[(CH3)3NH⁺] = ( 0.0005mol/25.00ml) × 1000ml = 0.0200M

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

pH = 9.87 + log( 0.0600M/0.0200M)

pH = 9.87 + 0.48

pH = 10.35

d)

At equivalence point

[(CH3)3NH⁺] = 0.050M

(CH3)3NH⁺ partly dissociate

(CH3)3NH⁺ + H2O <--------> (CH3)3N + H3O⁺

K_a = [(CH3)3N][H3O⁺] / [(CH3)3NH⁺] = 1.35×10^-10

at equilibrium

[(CH3)3NH⁺] = 0.05 - x

[(CH3)3N] = x

[H3O⁺] = x

x²/( 0.05 - x) = 1.35 ×10^-10

solving for x

x = 2.60 × 10^-6

[H3O⁺] = 2.60 ×10^-6M

pH = -log[H3O⁺]

pH = - log( 2.60 ×10^-6)

pH = 5.59

e)

Equivalence point = 20ml

Volume of HCl added = 25ml

excess volume of HCl added = 25ml - 20ml = 5ml

excess moles of HCl added = (0.100mol/1000ml) × 5ml = 0.0005mol

Total volume = 20ml + 25ml = 45ml

[H3O⁺] = ( 0.0005mol/45ml) ×1000ml = 0.01111M

pH = -log(0.01111M)

pH = 1.95