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 the following volumes of titrant have been added: (а) 0.0 mL (b) 6.0 mL (c) 12.0 mL pH pH pH (d) 18.0 mL (e) 24.0 mL (f) 45.6 mL рH - рH: = pH

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