An aqueous solution initially contains () and . The final concentration of in the solution is about — Ionic Equilibrium Chemistry Question
Question
An aqueous solution initially contains $0.01 \text{ M} – \text{RNH}_2$ ($K_b = 2.0 \times 10^{-6}$) and $10^{-4} \text{ M} – \text{NaOH}$. The final concentration of $\text{OH}^-$ in the solution is about
💡 Solution & Explanation
Let the concentration of $\text{OH}^-$ produced by $\text{RNH}_2$ be $x$. The total $[\text{OH}^-] = x + 10^{-4}$. $K_b = \frac{[\text{RNH}_3^+][\text{OH}^-]}{[\text{RNH}_2]} = \frac{x(x + 10^{-4})}{0.01} = 2.0 \times 10^{-6} \implies x^2 + 10^{-4}x - 2.0 \times 10^{-8} = 0$. Using the quadratic formula, $x = \frac{-10^{-4} + \sqrt{10^{-8} - 4(1)(-2.0 \times 10^{-8})}}{2} = \frac{-10^{-4} + \sqrt{9.0 \times 10^{-8}}}{2} = \frac{-10^{-4} + 3.0 \times 10^{-4}}{2} = 1.0 \times 10^{-4} \text{ M}$. The total $[\text{OH}^-] = x + 10^{-4} = 1.0 \times 10^{-4} + 1.0 \times 10^{-4} = 2.0 \times 10^{-4} \text{ M}$. Therefore, correct answer is B.