Question

Amphetamine (c9h13n) is a weak base with a pkb of 4.2. calculate the ph of a solution containing an amphetamine concentration of 225 mg>l

Answer 1

pH = 10.505

Step-by-step explanation:

Molar mass of Amphetamine ( C9H13N) = 135 g/mol

Given that the concentration of Amphetamine = 225 mg/L

mass of Amphetamine in one Liter =
`(225)/( 1000)` = 0.225 g

Number of moles of Amphetamine in one liter =
`\frac {0.225 g} { 135g/mol}`

= 0.001667 mol

∴ molarity = 0.0017 M

C₉H₁₃N + H₂O --------> C₉H₁₃NH⁺ + OH⁻

I(M) 0.001667 M 0 0

C(M) -x x x

E(M) 0.001667 - x x x

Pkb = -log Kb = 4.2

∴ Kb = 6.309 x 10⁻⁵

Kb = 6.309 x 10⁻⁵

Equilibrium constant = [C₉H₁₃NH⁺][OH⁻]/ [C₉H₁₃N]

6.309 x 10⁻⁵ = x² / 0.001667-x

where 0.001667 -x ≅ 0.001667

Then;

x² = 6.309 x 10⁻⁵ × 0.001667

x² = 1.0517103 × 10⁻⁷

x =
`\sqrt {1.0517103 * 10^(-7)`

x = 0.00032 M

x = [OH-] = 0.00032 M

∴ pOH = -log [OH-]

pOH = -log (0.00032)

pOH =3.495

pH = 14 - 3.495

= 10.505