Question

Nicotine, (C₁0H₁4N₂), is a base that will accept two protons ((K_b1 = 7 × 10^−7), (K_b2 = 1.4 × 10^−11)). What is the concentration of each species present in a 0.050-M solution of nicotine?

a) (H₂N⁻, OH⁻)
b) (H₂N⁻, H₂O)
c) (OH⁻, H₂O)
d) (H₂N⁻, H₃O⁺)

Answer 1

Nicotine, C₁₀H₁₄N₂, is a base that accepts two protons (K_b1 = 7 x 10-7, K_b2 = 1.4 x 10-₁¹). The concentration of each species present in a 0.050-M solution of nicotine can be calculated using the provided ionization constants.

Step-by-step explanation:

Nicotine, C₁₀H₁₄N₂, is a base that accepts two protons.

The equation for its dissociation is:

C₁₀H₁₄N₂ + H₂O ⇄ C₁₀H₁₄N⁺ + OH⁻

The concentration of each species present in a 0.050-M solution of nicotine can be calculated using the ionization constants (K_b1 and K_b2) provided:

Let x be the concentration of C₁₀H₁₄N²⁺.

From K_b1: K_b1 = [C₁₀H₁₄N⁺][OH⁻]/[C₁₀H₁₄N₂]
7 × 10⁻⁷ = x * x/0.050

From K_b2: K_b2 = [C₁₀H₁₄N²⁺][OH⁻]/[C₁₀H₁₄N⁺]
1.4 × 10⁻¹¹ = (x * x)/0.050

Solving these two equations simultaneously, we can find the concentration of each species present in the solution.