The equilibrium constant for the ionization of in water as<br><br><br><br>is at . Which of the follo — Ionic Equilibrium Chemistry Question
Question
The equilibrium constant for the ionization of $\text{RNH}_2(g)$ in water as<br><br>$\text{RNH}_2(g) + \text{H}_2\text{O}(l) \rightleftharpoons \text{RNH}_3^+(aq) + \text{OH}^-(aq)$<br><br>is $10^{-6}$ at $25^\circ\text{C}$. Which of the following is/are correct?
💡 Solution & Explanation
$K = \frac{[\text{RNH}_3^+][\text{OH}^-]}{P_{\text{RNH}_2}} = 10^{-6}$. If $P_{\text{RNH}_2} = 1\text{ bar}$, then $[\text{OH}^-]^2 = 10^{-6} \implies [\text{OH}^-] = 10^{-3}\text{ M}$, yielding $\text{pH} = 11.0$ (Option A). Adding HCl consumes $\text{OH}^-$, shifting equilibrium forward by Le Chatelier's principle (Option B). Adding $\text{H}_2\text{O}(l)$ expands the volume, decreasing concentrations of aqueous products, which makes the reaction quotient $Q < K$, shifting the reaction forward (Option C). Adding more $\text{RNH}_2(g)$ increases reactant pressure, shifting the equilibrium forward (Option D). Therefore, correct answer is A,B,C,D.