Question

Consider the titration of 40.0 mL of 0.0600 M (CH3)2NH (a weak base; Kb=0.000540) with 0.100M HIO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 ml (b) 6,0 ml. (c) 12.0 mL   (d) 18.0 mL (e) 24.0 ml (f) 38.4ml

Answer 1

HClO₄ is a strong acid.

(a) NH₃ + H₂O -------> NH₄⁺ +OH^-

I 0.06 M   

C -x   +x +x

E 0.06 - x x x

At 0.0 ml no HClO₄ is present

K_b = [ NH₄⁺ ][OH^-] / [NH₃]

1.8 x 10^-5 = ( x . x ) / (0.06 - x)

x² = 0.06 x 1.8 x 10^-5

x = 1.04 x 10^-3 M

pOH = -log[OH^-]

- log ( 1.04 x 10^-3 M)

pOH = 2.98

pH + pOH = 14

pH = 14 - pOH

=14 - 2.98

pH = 11.02

(b) 6.0 ml

moles of NH₃ = concentration x volume

= 0.06 M x 40 mL

= 2.4 mmol

moles of HClO₄ added = 0.1 M x 6.0 ml

= 0.6 mmol

   NH₃ + HClO₄ ------> NH₄⁺

I 2.4 mmol 0.6mmol --

C -0.6 mmol -0.6mmol +0.6mmol

E 1.8 mmol 0 0.6 mmol

Total volume = 40 mL + 6 mL = 46 mL

To calculate pH we have to usee Henderson Hasselbalch equation

pOH = pK_b + log ([NH₄⁺] / [NH₃])

pK_b = - log K_b = -log (1.8 x 10^-5 ) = 4.74

pOH = 4.74 + log ((0.6 x 46)/(46 x 1.8))

= 4.74 - 0.48

= 4.26

pH = 14- 4.26

pH = 9.74

(c) 12 mL

moles of HClO₄ added = 0.1 M x 12.0 ml

= 1.2 mmol

   NH₃ + HClO₄ ------> NH₄⁺

I 2.4 mmol 1.2 mmol --

C -1.2 mmol -1.2 mmol +1.2 mmol

E 1.2 mmol 0 1.2 mmol

Total volume = 40 mL + 12 mL = 52 mL

To calculate pH we have to usee Henderson Hasselbalch equation

pOH = pK_b + log ([NH₄⁺] / [NH₃])

pK_b = - log K_b = -log (1.8 x 10^-5 ) = 4.74

pOH = 4.74 + log ((1.2 x 52)/(52 x 1.2))

= 4.74 + 0

= 4.74

pH = 14- 4.74

pH = 9.26

(d) 18 mL

moles of HClO₄ added = 0.1 M x 18.0 ml

= 1.8 mmol

   NH₃ + HClO₄ ------> NH₄⁺

I 2.4 mmol 1.8 mmol --

C -1.8 mmol -1.8mmol +1.8mmol

E 0.6 mmol 0 1.8 mmol

Total volume = 40 mL + 18 mL = 58 mL

To calculate pH we have to usee Henderson Hasselbalch equation

pOH = pK_b + log ([NH₄⁺] / [NH₃])

pK_b = - log K_b = -log (1.8 x 10^-5 ) = 4.74

pOH = 4.74 + log ((1.8 x 58)/(58 x 0.6))

= 4.74 + 0.48

= 5.22

pH = 14- 5.22

pH = 8.78

(e) 24 mL

moles of HClO₄ added = 0.1 M x 24.0 ml

= 2.4 mmol

this is the equivalence point only NH₄⁺ is present

Total volume = 40 mL + 24 mL = 64 mL

concentration of [NH₄⁺] = mmol / total volume

= 2.4 mmol / 64 mL

  [NH₄⁺] = 0.0375 M

NH₄⁺ + H₂O ----------> NH₃ + H₃O⁺

I 0.0375 M --- ---

C -x +x +x

E 0.0375 - x x x

K_a = [NH₃] [H₃O⁺] / [NH₄⁺]

K_a of NH₄⁺ = K_w / K_b = 10^-14 / (1.8 x 10^-5) = 5.56 x 10^-10

5.56 x 10^-10 = x² / (0.0375 - x)

it's a quadratic equation solving it the correct x = 4.6 x 10^-6 M

[H₃O⁺] = 4.6 x 10^-6 M

pH = -log[4.6 x 10^-6 M]

pH = 5.34

(f) 38.4 mL

moles of HClO₄ added = 0.1 M x 38.4 ml

= 3.84 mmol

2.4 mmol NH3 reacts with 2.4 mmol of HClO4

excess HCLo4 = 3.84 - 2.4 = 1.44 mmol

total volume = 40 + 38.4 = 78.4 ml

concentration of excess [H+] = mmol / total volume

= 1.44 / 78.4 = 0.018 M

[H⁺] = 0.018 M

pH = -log[H⁺]

= - log (0.018)

pH = 1.74