An aqueous solution of aniline of concentration 0.2 M is prepared. How many milligrams of NaOH shoul — Ionic Equilibrium Chemistry Question
Question
An aqueous solution of aniline of concentration 0.2 M is prepared. How many milligrams of NaOH should be added in 500 ml of this solution so that anilinium ion concentration in the solution becomes $10^{-8}\text{ M}$? $K_b$ of $\text{C}_6\text{H}_5\text{NH}_2 = 4.0 \times 10^{-10}$.
💡 Solution & Explanation
The base dissociation is $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$. We require $[\text{BH}^+] = 10^{-8}\text{ M}$. The initial concentration of aniline is $[\text{B}] \approx 0.2\text{ M}$. Using $K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \implies 4.0 \times 10^{-10} = \frac{(10^{-8})[\text{OH}^-]}{0.2}$. Solving for hydroxide, $[\text{OH}^-] = \frac{4.0 \times 10^{-10} \times 0.2}{10^{-8}} = 8.0 \times 10^{-3}\text{ M}$. The moles of $\text{NaOH}$ required in $500\text{ ml}$ ($0.5\text{ L}$) is $8.0 \times 10^{-3} \times 0.5 = 4.0 \times 10^{-3}\text{ moles}$. The mass of $\text{NaOH}$ is $4.0 \times 10^{-3} \times 40\text{ g/mol} = 0.16\text{ g} = 160\text{ mg}$. Formatted to four digits, this is 0160.