Question

10a) A 29.7 mL sample of 0.275 M methylamine, CH3NH2, is titrated with 0.267 M nitric acid. At the equivalence point, the pH is _______.

10B)A 20.4 mL sample of 0.357 M trimethylamine, (CH₃)₃N, is titrated with 0.395 M hydrochloric acid.

After adding 28.0 mL of hydrochloric acid, the pH is____.

Answer 1

10 a )

Given data ,

M1 = 29.7 mL , V1 = 0.275 M , M2 = 0.267 M

At equivalence point, M1V1 = M2V2

  29.7 x 0.275 = 0.267 x V2

V2 = 8.16 / 0.267

V2 = 30.56 M

The concentration of salt formed at equivalence = 29.7 x 0.275 / (29.7 + 30.56 )

= 8.16 / 60.26

= 0.135 M

It is a salt of weak base and strong acid whose pH is given by

pH = 1/2 [ pKw -pkb - log C]

= 1/2 [ 14 - 3.36 - log 0.135]

pH = 9.78

10 b)

The reaction between (CH3)3N and HCl is

(CH3)3N + HCl - - - - - - > (CH3)3NH+ + Cl^-

1:1 molar reaction

The Initial moles of (CH3)3N = (0.357 mol/1000ml)× 20.4 ml

= 0.00728

No of moles of HCl added = (0.395 mol /1000ml) × 28ml

= 0.01106 mol

0.01106 moles of HCl react with 0.01106 moles of (CH3)3N to produce 0.01106moles of (CH3)3NH+

After adding HCl

Volume of solution = 20.4 mL + 28.0 mL

= 48.4 ml

[(CH3)3NH+] = (0.01106mol / 48.4 ml) × 1000ml

= 0.2285M

[(CH3)3N] = ((0.00728 mol- 0.01106mol) / 48.4 ml )×1000ml

= - 0.078099 M

pKa of (CH3)3NH+ = 9.81

Henderson-Hasselbalch equation is

pH = pKa + log([A^-] /[HA])

pH = 9.81 + log([(CH3)3N)]/[(CH3)3NH+])

pH = 9.81 + log(- 0.078099 M / 0.2285M )

pH = 9.81 + 0.46

pH = 10.27