Chemistry

Chemistry

Problem

Calculate the [OH−][OH^-] of [NH2⋅C2H4NH3]+[NH_2 \cdot C_2H_4NH_3]^+ and [H3N−C2H4NH3]2+[H_3N - C_2H_4NH_3]^{2+} in 0.15 M ethylene diamine (aq.); if NH2C2H4NH2+H2O⇌NH2C2H4NH3++OH−NH_2C_2H_4NH_2 + H_2O \rightleftharpoons NH_2C_2H_4NH_3^+ + OH^- K1=8.5×10−5K_1 = 8.5 \times 10^{-5} NH2C2H4NH3++H2O⇌[NH3C2H4NH3]2++OH−NH_2C_2H_4NH_3^+ + H_2O \rightleftharpoons [NH_3C_2H_4NH_3]^{2+} + OH^- K2=2.7×10−8K_2 = 2.7 \times 10^{-8}
Intuitive explanation

Ethylene diamine is a diprotic base with two nitrogen lone pairs, so it accepts protons in two sequential steps. The key is recognizing that the first ionization is much stronger (K1≫K2K_1 \gg K_2), meaning the first equilibrium dominates [OH−][OH^-] production, while the second step contributes only a tiny additional amount. Set up an ICE table for the first reaction, solve for [OH−][OH^-] and the intermediate species, then use that result to find how much further ionization occurs in the second step.

Intuitive explanation

Ethylene diamine is a diprotic base with two nitrogen lone pairs, so it accepts protons in two sequential steps. The key is recognizing that the first ionization is much stronger (K1≫K2K_1 \gg K_2), meaning the first equilibrium dominates [OH−][OH^-] production, while the second step contributes only a tiny additional amount. Set up an ICE table for the first reaction, solve for [OH−][OH^-] and the intermediate species, then use that result to find how much further ionization occurs in the second step.

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