Question

Amphetamine (C9H13N) is a weak base with a pKb of 4.2.

Calculate the pH of a solution containing an amphetamine concentration of 230 mg/L .

Answer 1

Given,

pK_b of a weak base Amphetamine(C₉H₁₃N) = 4.2

Also given,

the concentration of Amphetamine = 230 mg /L

Firstly converting this concentration to mol/L,

= 230 mg /L x ( 1 g / 1000 mg) x ( 1 mol / 135.206 g)

= 0.001701 mol/L

Now,

The equilbrium reaction for the weak base is,

C₉H₁₃N(aq) + H₂O(l) C₉H₁₃NH⁺(aq) + OH^-(aq)

Drawing an ICE chart,

	C9H13N(aq)	C9H13NH+(aq)	OH-(aq)
I(M)	0.001701	0	0
C(M)	-x	+x	+x
E(M)	0.001701-x	x	x

Now, Calculating K_b from the given pK_b value,

pK_b = -log K_b

4.2 = -log K_b

K_b = 6.3096 x 10^-5

Now, The K_b expression for the equilibrium reaction is,

K_b = [C₉H₁₃NH⁺][OH^-] / [C₉H₁₃N]

6.3096 x 10^-5 = x² / (0.001701-x)

6.3096 x 10^-5 = x² / (0.001701) --------------- Here,(0.001701-x) 0.001701 Since, x<<<0.001701

x = 0.0003276

Thus, [OH^-] = x = 0.0003276 M

We know,

[H₃O⁺] [OH^-] = 1.0 x 10^-14

[H₃O⁺] [0.0003276] = 1.0 x 10^-14

[H₃O⁺] = 3.0523 x 10^-11 M

WE know, the formula to calculate pH,

pH = -log [H₃O⁺]

pH = -log [3.0523 x 10^-11]

pH = 10.5