Question

1)A 21.7 mL sample of 0.267 M dimethylamine, (CH3)2NH, is titrated with 0.241 M perchloric acid. The pH before the addition of any perchloric acid is

2)A 27.1 mL sample of 0.212 M dimethylamine, (CH3)2NH, is titrated with 0.201 M hydroiodic acid. After adding 40.6 mL of hydroiodic acid, the pH is

Answer 1

dimethyl amine has pkb of 3.29 , Kb = 10 ^- 3.29 = 0.000513

(CH3)2NH moles = M x V ( in L) = 0.267 x 0.0217 = 0.005794

The base dissociation eq is given by

               (CH3)2NH (aq) + H2O (l)   <----> (CH3)2NH2+ (aq) +    OH- (aq)

Initial            0.005794                                            0                      0

equilibrium    0.005794-X                                         X                        X

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

0.000513 = ( X/0.0217)^2 / ( 0.005794-X) / (0.0217)

X^2 / ( 0.005794-X) = 0.0000111

X^2 + 0.0000111X - 6.45 x 10^-8 = 0

X = 0.0002485 , [OH-] = 0.0002485 /0.0217 = 0.01145

pOH = -log [OH-] = -log ( 0.01145) = 1.94

pH = 14-pOH = 14-1.94 = 12.06

2) HI acid moles added = M x V = 0.201 x 0.0406 = 0.00816

(CH3)2NH moles = 0.212 x 0.0217 = 0.0046

we had excess of acid compared to base

Acid and base react in 1:1 ratio

excess acid moles = 0.00816 -0.0046 = 0.00356

total solution volume = 27.1 + 40.6 = 67.7 ml = 0.0677 L

[HI] = moles / volume = 0.00356 / 0.0677 = 0.0526

[H+] = 0.0526       ( since 1HI gives 1H+ )

pH = -log [H+] = -log (0.0526) = 1.28