Question

A 100.0-mL sample of 0.2 M (CH3)3N (Kb = 5.28 × 10–5) is titrated with 0.2 M HCl. What is the pH at the equivalence point?

Answer 1

Given,

Concentration of (CH₃)₃N = 0.2 M

Volume of (CH₃)₃N sample = 100.0 mL x ( 1L/1000 mL) = 0.1L

Concentration of HCl solution = 0.2 M

Now,

Firstly calculating the number of moles of  (CH₃)₃N,

= 0.2 M x 0.1 L

= 0.02 moles of  (CH₃)₃N

Now, We know, At the equivalence point,

Moles of Acid = Moles of base

Thus, Moles of HCl = Moles of  (CH₃)₃N

Moles of HCl = 0.02 moles

Now, Calculating the volume of HCl solution required from the calculated moles and the given concentration,

= 0.02 moles of HCl / 0.2 M

= 0.1 L of HCl solution

Thus, the total volume = 0.1 L of  (CH₃)₃N + 0.1 L HCl = 0.2 L of solution at equivalence point

Thus,

(CH₃)₃N(aq) + HCl   (CH₃)₃NH⁺(aq)

	(CH3)3N(aq)	HCl	(CH3)3NH+(aq)
I(moles)	0.02	0.02	0
C(moles)	-0.02	-0.02	+0.02
E(moles)	0	0	0.02

Thus, [ (CH₃)₃NH⁺] = 0.02 moles /0.2 L solution

[(CH₃)₃NH⁺] = 0.1 M

Now, The equilbrium reaction for  (CH₃)₃NH⁺ is,

(CH₃)₃NH⁺(aq) + H₂O(l)   (CH₃)₃N(aq) + H₃O⁺(aq)

Drawing an ICE chart for the above equilibrium,

	(CH3)3NH+(aq)	(CH3)3N(aq)	H3O+(aq)
I(M)	0.1	0	0
C(M)	-x	+x	+x
E(M)	0.1-x	x	x

K_a x K_b = K_w

K_a x (5.28 x 10^-5) = 1.0 x 10^-14

K_a = 1.894 x 10^-10

Now, the K_a expression is,

K_a = [(CH₃)₃N] [H₃O⁺] / [(CH₃)₃NH⁺]

Substituting the known values,

1.894 x 10^-10 = x² / (0.1-x)

1.894 x 10^-10 = x² / (x) -----------Here, (0.1-x) 0.1 Since, x<<<< 0.1

x = 4.352 x 10^-6

Now, [H₃O⁺] = x = 4.352 x 10^-6

We know,

pH = -log[H₃O⁺]

pH = -log[4.352 x 10^-6 ]

pH = 5.4