Liquid ammonia ionizes to slight extent. At , its self-ionization constant, . How many amide ions ar — Ionic Equilibrium Chemistry Question
Question
Liquid ammonia ionizes to slight extent. At $-50^\circ \text{C}$, its self-ionization constant, $K = [\text{NH}_4^+][\text{NH}_2^-] = 10^{-30} \text{ M}^2$. How many amide ions are present per ml of pure liquid ammonia? ($N_A = 6 \times 10^{23}$)
Answer: C
💡 Solution & Explanation
In pure liquid ammonia, $[\text{NH}_4^+] = [\text{NH}_2^-] = \sqrt{10^{-30}} = 10^{-15} \text{ M}$. The number of moles per ml is $10^{-15} \text{ mol/L} \times 10^{-3} \text{ L/ml} = 10^{-18} \text{ mol/ml}$. The number of amide ions per ml is $10^{-18} \times 6 \times 10^{23} = 6 \times 10^5$ ions. Therefore, correct answer is C.
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