Question

A tablet of a common over-the-counter drug contains 400. mg of caffeine (C₈H₁₀N₄O₂) (MM=194.1906 g/mol). What is the pH of the solution resulting from the dissolution of a tablet in 225. mL of water? (For caffeine, K_b = 4.1 × 10^–4.)

Answer choices:

7.67

2.76

11.00

11.29

6.33

Answer 1

Answer) 11.29

Explanation:

1g = 1000 mg

Mass of caffeine = 400mg/1000mg = 0.4g

1 L = 1000 mL

Volume of solution = 225 mL /1000 mL = 0.225 L

Number of moles of caffeine = mass÷molar mass

Number of moles = 0.4g÷194.1906g/mol =0.0020598 mol

Concentration of caffeine = number of moles÷volume of solution in litres
Concentration of caffeine = 0.0020598 mol/0.225L = 0.00915M of caffeine

Write the base dissociation equilibrium reaction:
C₈H₁₀N₄O₂ + H₂O C₈H₁₀N₄O₂H⁺ + OH^-

Initial. 0.00915 M 0 0
Change. -x + x +x
Equilibrium 0.00915 - x x x

Write the expression for base dissociation constant:

K_b = [C₈H₁₀N₄O₂H⁺][OH^-]/[C₈H₁₀N₄O₂]
K_b = x²/ 0.00915 - x

Caffeine is a weak base, therefore dissociation is less. Hence, 0.00915 - x ~ 0.00915

K_b = x²/ 0.00915

( 4.1 x 10^-4 × 0.00915)^1/2 = x

x = 1.936 x10^-3 M = [OH^-]

Calculate pOH of solution using the equation as follows:
pOH = -log[OH^-]

pOH = - log(1.936 x10^-3 M) = 2.71

Calculate pH of solution using the equation as follows:
pH= 14 - pOH

pH = 14 - 2.71 = 11.29

Hence, the pH of the pH of the solution is 11.29