Question

Consider the titration of 30.0 mL of 0.0700 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HClO4. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL

pH =

(b) 5.3 mL

pH =

(c) 10.5 mL

pH =

(d) 15.8 mL

pH =

(e) 21.0 mL

pH =

(f) 33.6 mL

pH =

Answer 1

Kb of (CH₃)₂NH = 0.000540

(CH₃)₂NH + H₂O -----> (CH₃)₂NH₂⁺ + OH^-

Kb = [(CH₃)₂NH₂⁺][ OH^-] / [(CH₃)₂NH] = 0.000540

(a)

Let s = amount of (CH₃)₂NH that dissociates (ionizes).

[(CH₃)₂NH₂⁺] = s

[OH^-] = s

[(CH₃)₂NH] = 0.070 – s

5.4 x 10^-4 = (s)(s) / (0.07 - s)

5.4 x 10^-4 = s² / 0.07 (assuming that s << 0.50)

s² = (5.4 x 10^-4 x 0.07) = 3.78 x 10^-5

s = 6.15 x 10^-3

[OH^-] = s =6.15 x 10^-3

pOH = - log (6.15 x 10^-3) = 2.21

pH + pOH = 14

pH = 14 - pOH = 14 – 2.21 = 11.79

(b)

HClO₄ reacts with (CH₃)₂NH to form (CH₃)₂NH₂⁺

moles (CH₃)₂NH initially = 0.07 M x 0.030 L = 0.0021 mol (CH₃)₂NH

moles HClO₄ added = 0.1 M x 0.0053 L = 0.00053 mol HClO₄ added

moles (CH₃)₂NH₂⁺ formed = 0.00053 mol

moles (CH₃)₂NH remaining = (0.0021 -0.00053) mol = 0.00157

K_a = 1 x 10^-14 / K_b

    = 1 x 10^-14 / 5.4 x 10^-4 = 1.85 x 10^-11

pKa = -log Ka = -log (1.85 x 10^-11) = 10.73

Using the Henderson-Haselbalch equation.

pH = pKa + log [(CH₃)₂NH₂⁺]/[(CH₃)₂NH]

    = 10.73 + log [0.00053/0.00157]

    = 10.73 -0.47

    = 10.26

(c)

HClO₄ reacts with (CH₃)₂NH to form (CH₃)₂NH₂⁺

moles (CH₃)₂NH initially = 0.07 M x 0.030 L = 0.0021 mol (CH₃)₂NH

moles HClO₄ added = 0.1 M x 0.0105 L = 0.00105 mol HClO₄ added

moles (CH₃)₂NH₂⁺ formed = 0.00105 mol

moles (CH₃)₂NH remaining = (0.0021 -0.00105) mol = 0.00105

K_a = 1 x 10^-14 / K_b

    = 1 x 10^-14 / 5.4 x 10^-4 = 1.85 x 10^-11

pKa = -log Ka = -log (1.85 x 10^-11) = 10.73

Using the Henderson-Haselbalch equation.

pH = pKa + log [(CH₃)₂NH₂⁺]/[(CH₃)₂NH]

    = 10.73 + log [0.00105/0.00105]

    = 10.73

    = 10.26

(d)

HClO₄ reacts with (CH₃)₂NH to form (CH₃)₂NH₂⁺

moles (CH₃)₂NH initially = 0.07 M x 0.030 L = 0.0021 mol (CH₃)₂NH

moles HClO₄ added = 0.1 M x 0.0158 L = 0.00158 mol HClO₄ added

moles (CH₃)₂NH₂⁺ formed = 0.00158 mol

moles (CH₃)₂NH remaining = (0.0021 -0.00158) mol = 0.00052

K_a = 1 x 10^-14 / K_b

    = 1 x 10^-14 / 5.4 x 10^-4 = 1.85 x 10^-11

pKa = -log Ka = -log (1.85 x 10^-11) = 10.73

Using the Henderson-Haselbalch equation.

pH = pKa + log [(CH₃)₂NH₂⁺]/[(CH₃)₂NH]

    = 10.73 + log [0.00158/0.00052]

    = 10.73 +0.48

    = 11.21

(e)

HClO₄ reacts with (CH₃)₂NH to form (CH₃)₂NH₂⁺

moles (CH₃)₂NH initially = 0.07 M x 0.030 L = 0.0021 mol (CH₃)₂NH

moles HClO₄ added = 0.1 M x 0.021 L = 0.0021 mol HClO₄ added

moles (CH₃)₂NH₂⁺ formed = 0.0021 mol

[(CH₃)₂NH₂⁺] = 0.0021 mol / (30 + 21) mL

                      = 0.0021 mol / 0.051 L = 0.041 M

So, this is the equivalence point, the solution contains ONLY (CH₃)₂NH₂⁺

(CH₃)₂NH₂⁺ = (CH₃)₂NH + H⁺

Ka = [(CH₃)₂NH] [H⁺] / [(CH₃)₂NH₂⁺]

1.85 x 10^-11 = x² / 0.041

x² = 7.6 x 10^-13

x = 8.72 x 10^-7

[(CH₃)₂NH] = [H⁺] = x = 8.72 x 10^-7

pH = -log [H⁺] = -log (8.72 x 10^-7) = 6.06

(f)

HClO₄ reacts with (CH₃)₂NH to form (CH₃)₂NH₂⁺

moles (CH₃)₂NH initially = 0.07 M x 0.030 L = 0.0021 mol (CH₃)₂NH

moles HClO₄ added = 0.1 M x 0.0336 L = 0.00336 mol HClO₄ added

Excess moles HClO₄ = 0.00126 mol

[HClO₄] = 0.00126 mol / 0.0636 L = 0.02 M

Since HClO₄ is a strong, it will dissociate completely.

So, [H⁺] = 0.02 M

pH = -log[H⁺] = -log (0.02) = 1.7