Question

The pH of an aqueous solution of 0.456 M trimethylamine (a weak base with the formula (CH3)3N) is

Answer 1

The ionization of  (CH3)3N) in water is represented as

(CH₃)₃N (aq)+ H2O(l) → (CH₃)₃NH⁺(aq) + OH^– (aq)

As  (CH₃)₃N is a weak base and ionizes to a lesser extent, we must use K_b value to determine equilibrium concentration of OH^–

Setting up ICE table to determine equilibrium concentration of all species

	(CH3)3N (aq)	(CH3)3NH+(aq)	OH– (aq)
Initial conc	0.456 M	0	0
change in conc	-x M	+x M	+x M
Equilibrium conc	(0.456-x) M	x M	x M

K_b of  (CH₃)₃N (aq) = [(CH₃)₃NH⁺(aq) ][OH^– (aq)] / [(CH₃)₃N (aq)]

= x.x / (0.456-x)

=x²/(0.456-x)

Assuming x to be small and ignoring it from denominator

K_b =x²/0.456

K_b of  (CH₃)₃N (aq) = 6.4 *10^-5 -----[literature value]

6.4 *10^-5 = x²/0.456

x = 0.0054

Thererefore, at equilibrium [(CH₃)₃NH⁺(aq) ]=[OH^– (aq)] = 0.0054 M

pOH = -log[ OH^– ] = -log(0.0054)

=2.27

pH+pOH= 14

pH= 14- pOH

= 14 - 2.27

=11.73

pH of 0.456 M CH₃)₃N = 11.73