Question

A 21.5 mL sample of 0.383 M dimethylamine, (CH32NH, is titrated with 0.279 M perchloric acid. (1) Before the addition of any perchloric acid, the pH is (2) After adding 11.8 mL of perchloric acid, the pH is (3) At the titration midpoint, the pH is (4) At the equivalence point, the pH is (5) After adding 42.2 mL of perchloric acid, the pH is

Ka of dimethylamine = 5.9*10^-4

Answer 1

(a)

(CH₃)₂NH is a weak base.

Ka of (CH₃)₂NH = 5.9 * 10^-4

So, Kb = Kw / Ka
            = (1.0 * 10^-14) / (5.9 * 10^-4)
            = 1.7 * 10^-11

Now,

                 (CH₃)₂NH    +    H2O      <------->        (CH₃)₂NH₂⁺     +     OH^-
IC:               0.383                                                        0                      0
C:                 - x                                                          + x                    + x
EC:          0.383 – x                                                      x                      x

Kb = [(CH₃)₂NH₂⁺] [OH^-] / [(CH₃)₂NH]
or, 1.7 * 10^-11 = (x) (x) / (0.383 – x)
or, 1.7 * 10^-11 = x² / (0.383 – x)
or, 1.7 * 10^-11 = x² / (0.383)          [Kb is very small, so x term in denominator is discarded]
or, x² = (1.7 * 10^-11) * (0.383)
or, x² = 6.5 * 10^-12
or, x = 2.55 * 10^-6

So, [OH-] = x = 2.55 * 10^-6

pOH = - log [OH-]
        = - log (2.55 * 10^-6)
        = 5.59

Now, pH = 14 – pOH
              = 14 – 5.59
              = 8.41

(2)

Moles of (CH₃)₂NH = 0.383 M x 0.0215 L = 0.0082345 mol
Moles of perchloric acid = 0.279 M x 0.0118 L = 0.0032922 mol

So, 0.0032922 mol of perchloric acid will react with 0.0032922 mol of (CH₃)₂NH to produce 0.0032922 mol of (CH₃)₂NH₂⁺.

So, mole of (CH₃)₂NH left = 0.0082345 mol - 0.0032922 mol = 0.0049423 mol

Total volume = 21.5 mL + 11.8 mL = 33.3 mL = 0.033 L

So, the concentration terms are
[(CH₃)₂NH] = 0.0049423 mol / 0.033 L = 0.1498 M
[(CH₃)₂NH₂⁺] = 0.0032922 mol / 0.033 L = 0.0998 M

Using the Henderson - Hasselbalch equation

pOH = pKb + log ([salt] / [base])
pOH = pKb + log { [(CH₃)₂NH₂⁺] / [(CH₃)₂NH] }

Some preliminary calculations :
Kb = 1.7 * 10^-11

So, = pKb = - log Kb = - log (1.7*10^-11) = 10.77

Now,

pOH = pKb + log { [CH₃)₂NH₂⁺] / [CH₃)₂NH] }
pOH = 10.77 + log (0.0998 / 0.1498)
pOH = 10.77 + log (0.666)
pOH = 10.77 - 0.18
pOH = 10.59

Now, pH = 14 – pOH
              = 14 – 10.59
              = 3.41

(3)

At mid point;
Half of the [(CH₃)₂NH is neutralized by the acid. Therefore, the moles of (CH₃)₂NH = moles of (CH₃)₂NH₂⁺.

So, [(CH₃)₂NH] = [(CH₃)₂NH₂⁺]

Using the Henderson - Hasselbalch equation

pOH = pKb + log { [(CH₃)₂NH₂⁺] / [(CH₃)₂NH] }
pOH = pKb + log 1
pOH = 10.77 +
pOH = 10.77

Now, pH = 14 – pOH
              = 14 – 10.77
              = 3.23

4.

At equivalence point;

At equivalence point, all the base [(CH₃)₂NH] is neutralized with acid.

Moles of (CH₃)₂NH = 0.0082345 mol

So, moles of perchloric acid = 0.0082345 mol
molarity of perchloric acid = 0.279 M
So, volume of perchloric acid = 0.0082345 mol / 0.279 M = 0.0295 L = 29.5 mL

Total volume = 21.5 mL + 29.5 mL = 51 mL = 0.051 L

Now, 0.0082345 mol of perchloric acid will react with 0.0082345 mol of (CH₃)₂NH to produce 0.0082345 mol of (CH₃)₂NH₂⁺.

[(CH₃)₂NH₂⁺] = (0.0082345 mol) / (0.051 L) = 0.16 M

Now,

                 (CH₃)₂NH₂⁺          <------->         (CH₃)₂NH       +       H⁺
IC:               0.16                                               0                      0
C:                 - x                                               + x                    + x
EC:          0.16 – x                                              x                      x

Ka = [(CH₃)₂NH] [H⁺] / [(CH₃)₂NH₂⁺]
or, 5.9*10^-4 = (x) (x) / (0.16 – x)
or, 5.9*10^-4 = x² / (0.16 – x)
or, 5.9*10^-4 = x² / (0.16)          [Ka is very small, so x term in denominator is discarded]
or, x² = (5.9*10^-4) * (0.16)
or, x² = 9.44 * 10^-5
or, x = 0.0097

So, [H⁺] = x = 0.0097

pH = - log [H⁺]
        = - log (0.0097)
        = 2.01

(5)

Moles of (CH₃)₂NH = 0.383 M x 0.0215 L = 0.0082345 mol
Moles of perchloric acid = 0.279 M x 0.0422 L = 0.0117738 mol

So, 0.0082345 mol of CH₃)₂NH will react with 0.0082345 mol of perchloric acid.

So, mole of perchloric acid left = 0.0117738 mol - 0.0082345 mol = 0.0035393 mol

Total volume = 21.5 mL + 42.2 mL = 63.7 mL = 0.0637 L

So, [perchloric acid] = 0.0035393 mol / 0.0637 L = 0.056 M

Perchloric acid is a strong acid and hence will dissociate completely.

So, [H⁺] = 0.056 M

pH = - log [H⁺]
    = - log (0.056)
      = 1.25