Question

A 25.9 mL sample of 0.207 M trimethylamine, (CH3)2N, is titrated with 0.223 M nitric acid After adding 8.44 mL of nitric acid, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

1.Sol :-

Number of moles of (CH3)3N = Molarity of (CH3)3N x Volume in L

= 0.207 M x 0.0259 L

= 0.0053613 mol

Similarly,

Number of moles of HNO3 added = Molarity of HNO3​​​​​​​ x Volume in L

= 0.223 M x 0.00844 L

= 0.00188212 mol

ICF table is :

..........................(CH3)3N (aq).................+...............HNO3 (aq)-------------> (CH3)3NH+ (aq)..........+...........NO3- (aq)

Initial (I)................0.0053613 mol .............................0.00188212 mol..............0.0 mol.......................................

Change (C) .........-0.00188212 mol........................-0.00188212 mol..............+0.00188212 mol....................

Equilibrium (E).......0.00347918 mol...........................0.0 mol..............................0.00188212 mol..................

Using Henderson-Hasselbalch equation to determine the pH of basic buffer solution :

pOH = pKb + log [Conjugate acid] / [Base]

pOH = pKb of (CH3)3N + log [(CH3)3NH+] / [(CH3)3N]

= 4.19 + log 0.00188212 / 0.00347918

= 419 - 0.2668

= 3.92

As,

pH + pOH = 14

So,

pH = 14 - pOH

= 14 - 3.92

= 10.08

Hence, pH of the solution = 10.08