Question

Codeine (C18H21NO3, 299.4 g/mol) is a weak base. Suppose a 2.00 g pain tablet containing 30.0 mg codeine is dissolved in water to produce a 0.100 L solution with a pH of 9.470. Based on this information, determine the value of Kb for codeine.

Answer 1

Given,

Mass of codeine present in 2.00 g tablet = 30.0 mg

Volume of solution = 0.100 L

Molar mass of codeine = 299.4 g/mol

pH of codeine solution = 9.470

Calculating the number of moles of codeine from the given mass and molar mass,

= 30.0 mg x ( 1 g /1000 mg) x (1 mol / 299.4 g)

= 0.0001 mol codeine(C₁₈H₂₁NO₃)

Now, calculating the concentration of codeine solution,

We know, the formula to calculate molarity,

Molarity = Number of moles / L of solution

Substituting the values,

Molarity = 0.0001 mol / 0.1 L of solution

Molarity = 0.001 M

Now, calculating the [OH^-] from the given pH value,

pH = -log[H₃O⁺]

Rearranging the formula,

[H₃O⁺] = 10^-pH

[H₃O⁺] = 10^-9.470

[H₃O⁺] = 3.39 x 10^-10

Now, we know,

[H₃O⁺][OH^-] = 1.0 x 10^-14

Rearranging the formula,

[OH^-] = 1.0 x 10^-14/ [H₃O⁺]

[OH^-] = 1.0 x 10^-14/ 3.39 x 10^-10

[OH^-] = 2.95 x 10^-5 M

Now, the equilibrium reaction of codeine is,

C₁₈H₂₁NO₃(aq) + H₂O(l) HC₁₈H₂₁NO₃⁺(aq) + OH^-(aq)

Drawing an ICE chart.

	C18H21NO3(aq)	HC18H21NO3+(aq)	OH-(aq)
I(M)	0.001	0	0
C(M)	-x	+x	+x
E(M)	0.001-x	x	x

Now, we have,

[OH^-] = x = 2.95 x 10^-5 M

Calculating the equilibrium concentration values,

[C₁₈H₂₁NO₃]_eq = (0.001-x) = 0.00097 M

[HC₁₈H₂₁NO₃⁺]_eq = x = 2.95 x 10^-5 M

[OH^-]_eq = x = 2.95 x 10^-5 M

Now, the K_b expression is,

K_b = [HC₁₈H₂₁NO₃⁺][OH^-] /[C₁₈H₂₁NO₃]

K_b = [x][x] /[0.001-x]

K_b = [2.95 x 10^-5]² / [0.00097]

K_b = 8.96 x 10^-7