The only incorrect information related with solution of , ethylenediamine (en) is (, , , ) — Ionic Equilibrium Chemistry Question
Question
The only incorrect information related with $0.09 \text{ M}$ solution of $\text{NH}_2\text{CH}_2\text{CH}_2\text{NH}_2$, ethylenediamine (en) is ($K_{b1} = 8.1 \times 10^{-5}$, $K_{b2} = 7.0 \times 10^{-8}$, $\log 3 = 0.48$, $\log 7 = 0.85$)
💡 Solution & Explanation
Dissociation is dominated by $K_{b1}$. $[\text{OH}^-] \approx \sqrt{K_{b1} C} = \sqrt{8.1 \times 10^{-5} \times 0.09} = \sqrt{7.29 \times 10^{-6}} = 2.7 \times 10^{-3} \text{ M}$. Thus $[\text{enH}^+] \approx 2.7 \times 10^{-3} \text{ M}$ (Statement B is correct). For a dibasic system, the concentration of the doubly ionized species $[\text{enH}_2^{2+}] \approx K_{b2} = 7.0 \times 10^{-8} \text{ M}$ (Statement C is correct). $\text{pOH} = -\log(2.7 \times 10^{-3}) = 3 - \log 2.7 \approx 2.57$, so $\text{pH} = 14 - 2.57 = 11.43 \approx 11.44$ (Statement A is correct). The concentration of $[\text{H}^+] = 10^{-14} / 2.7 \times 10^{-3} = 3.7 \times 10^{-12} \text{ M}$, which is not $2.7 \times 10^{-3} \text{ M}$. Thus, D is incorrect. Therefore, correct answer is D.