Question

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

Calculate the pH of a solution containing an amphetamine concentration of 210 mg/L .

Answer 1

pKb = -log(Kb)

4.2 = -log (Kb)

Kb = 10^-4.2 = 6.3 x 10^-5 = [OH] * [C₉H₁₃NH⁺] / [C₉H₁₃NH⁺]

210 mg/L * (1g / 1000mg) = .21 g/L * (1mol/135.2g) = 0.00155 M

We now make our ICE table:

	C9H13N	<->	OH-	+	C9H13NH+
I	0.00155 M		0		0
C	-x		+x		+x
E	0.00155 - x		x		x

We now substitute:

6.3 x 10^-5 = x² / (0.00155 - x)

x = 0.000282574

This is the OH^- concentration

pOH = -log [OH^-]

pOH = - log [0.000282574]

pOH = 3.548868

pH = 14 - 3.548868 = 10.45 = pH