Question

Consider the titration of 40.0 mL of 0.0600 M (CH3)3N (a weak base; Kg = 6.40e-05) with 0.100 M HCl. Calculate the pH after t

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Answer #1

Ans:

No. of millimoles of (CH3)3N - weak base taken = 40 mL x 0.06M = 2.4 mmol

Kb of (CH3)3N = 6.4 x10-5

pKb of (CH3)3N = -logKb = -log (6.4 x10-5) = 4.19

Reaction: (CH3)3N + HCl ----> (CH3)3NHCl trimethyl ammonium chloride salt)

(CH3)3N

mmol

0.1M HCl added (mmol)

(CH3)3NHCl Salt formed (mmol)

Final solution consists of

a

2.4

0 mmol

0

2.4 mmol of (CH3)3N

b

6 mL x 0.1M = 0.6

0.6

1.8 mmol (CH3)3N + 0.6 mmol of (CH3)3NHCl

c

12 mL x 0.1M =1.2

1.2

1.2 mmol (CH3)3N + 1.2 mmol of (CH3)3NHCl

d

18 mL x 0.1M =1.8

1.8

0.6 mmol (CH3)3N + 1.8 mmol of (CH3)3NHCl

e

24 mL x 0.1M =2.4

2.4

2.4 mmol of (CH3)3NHCl

f

45.6 mL x 0.1M =4.56

2.4

2.4 mmol of (CH3)3NHCl + 2.16 mmol of HCl

Final solution consists of

pH is because of

pH formula

a

2.4 mmol of (CH3)3N

Weak base

pH = 14-pOH = 14-{-log[OH-]} = 14 + log(Kb.C)½

{C = molarity of (CH3)3N }

b

1.8 mmol (CH3)3N + 0.6 mmol of (CH3)3NHCl

Buffer

Henderson’s equation:

pOH = pKb + log ([Salt]/[base])

pOH = pKb + log ([(CH3)3NHCl]/[(CH3)3N])

pH = 14-pOH

c

1.2 mmol (CH3)3N + 1.2 mmol of (CH3)3NHCl

d

0.6 mmol (CH3)3N + 1.8 mmol of (CH3)3NHCl

e

2.4 mmol of (CH3)3NHCl

Salt made up of weak base & strong acid

pH = 7 – {(pKb + log C)/2}

C = molarity of salt

f

2.4 mmol of (CH3)3NHCl + 2.16 mmol of HCl

Acid HCl

(pH contribution from the salt can be ignored)

pH = -log[H+]

Final solution consists of

Final volume of the solution (mL)

Required concentrations

a

2.4 mmol of (CH3)3N

40

Molarity of (CH3)3N, C = 0.06 M

b

1.8 mmol (CH3)3N +

0.6 mmol of (CH3)3NHCl

46

[(CH3)3NHCl] = 0.6mmol/46mL = 0.013 M

[(CH3)3N] = 1.8mmol/46mL = 0.039 M

c

1.2 mmol (CH3)3N +

1.2 mmol of (CH3)3NHCl

52

[(CH3)3NHCl] = 1.2mmol/52mL = 0.023 M

[(CH3)3N] = 1.2mmol/52mL = 0.023 M

d

0.6 mmol (CH3)3N +

1.8 mmol of (CH3)3NHCl

58

[(CH3)3NHCl] = 1.8mmol/58mL = 0.031 M

[(CH3)3N] = 0.6mmol/58mL =0.010 M

e

2.4 mmol of (CH3)3NHCl

64

Molarity of salt, C = 2.4mmol/64mL =0.0375 M

f

2.4 mmol of (CH3)3NHCl + 2.16 mmol of HCl

85.6

Molarity of H+ = [H+] = 2.16mmol/85.6mL = 0.025 M

Required concentrations

pH formula

pH

a

Molarity of (CH3)3N, C = 0.06 M

pH = 14 + log(Kb.C)½

pH = 14 + log(6.4 x10-5 x 0.06)½

11.29

b

[(CH3)3NHCl] = 0.6mmol/46mL = 0.013 M

[(CH3)3N] = 1.8mmol/46mL = 0.039 M

pOH = pKb + log ([(CH3)3NHCl]/[(CH3)3N])

pOH = 4.19 + log(0.013/0.039)

pOH = 3.713

pH = 14-pOH = 10.287

10.287

c

[(CH3)3NHCl] = 1.2mmol/52mL = 0.023 M

[(CH3)3N] = 1.2mmol/52mL = 0.023 M

pOH = pKb + log ([(CH3)3NHCl]/[(CH3)3N])

pOH = 4.19 + log(0.023/0.023)

pOH = 4.19

pH = 14-4.19 = 9.81

9.81

d

[(CH3)3NHCl] = 1.8mmol/58mL = 0.031 M

[(CH3)3N] = 0.6mmol/58mL =0.010 M

pOH = pKb + log ([(CH3)3NHCl]/[(CH3)3N])

pOH = 4.19 + log(0.031/0.010)

pOH = 4.681

pH = 14- = 9.319

9.319

e

Molarity of salt, C = 2.4mmol/64mL =0.0375 M

pH = 7 – {(pKb + log C)/2}

pH = 7 – {(4.19 + log0.0375)/2} =

5.618

f

Molarity of H+ = [H+] = 2.16mmol/85.6mL =0.025 M

pH = -log[H+]

pH = -log0.025 =

1.602

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