Liquid self-ionizes to a slight extent. At a certain temperature, its self-ionization constant . Cal — Ionic Equilibrium Chemistry Question
Question
Liquid $NH_3$ self-ionizes to a slight extent. At a certain temperature, its self-ionization constant $K_{SIC} = 10^{-30}$. Calculate the number of $NH_4^+$ ions present per $100 \text{ cm}^3$ of pure liquid $NH_3$. The result is $a \times 10^7$ ions. Find the exact value of $a$ assuming Avogadro's number $N_A = 6.022 \times 10^{23}$.
💡 Solution & Explanation
The auto-ionization reaction is $2NH_3 \rightleftharpoons NH_4^+ + NH_2^-$. The self-ionization constant $K_{SIC} = [NH_4^+][NH_2^-] = 10^{-30}$. Assuming pure $NH_3$, $[NH_4^+] = \sqrt{10^{-30}} = 10^{-15} \text{ M}$. The volume is $100 \text{ cm}^3 = 0.1 \text{ L}$. Moles of $NH_4^+$ = $10^{-15} \times 0.1 = 10^{-16} \text{ mol}$. Number of ions = $10^{-16} \times (6.022 \times 10^{23}) = 6.022 \times 10^7$. Thus, $a = 6.022$.