A volume of of at 1 atm and was passed in of NaOH solution of unknown molarity. The resulting soluti — Ionic Equilibrium Chemistry Question
Question
A volume of $224\text{ ml}$ of $\text{CO}_2(\text{g})$ at 1 atm and $0^\circ\text{C}$ was passed in $1\text{ L}$ of NaOH solution of unknown molarity. The resulting solution when titrated with $1.0\text{ M} – \text{HCl}$ solution, requires $30\text{ ml}$ for the phenolphthalein end point. The molarity of NaOH solution used is
💡 Solution & Explanation
Moles of $\text{CO}_2 = 224\text{ ml} / 22400\text{ ml/mol} = 0.01\text{ moles}$. This reacts with NaOH to form $0.01\text{ moles}$ of $\text{Na}_2\text{CO}_3$. Let the initial moles of NaOH in $1\text{ L}$ be $x$. The unreacted NaOH is $(x - 0.02)\text{ moles}$. At the phenolphthalein end point, the HCl neutralizes all the free NaOH and converts $\text{Na}_2\text{CO}_3$ to $\text{NaHCO}_3$. The milliequivalents of HCl used = $30\text{ ml} \times 1.0\text{ M} = 30\text{ mmol} = 0.03\text{ moles}$. The equation for titration is: $n_{\text{HCl}} = n_{\text{NaOH(excess)}} + n_{\text{Na}_2\text{CO}_3} \implies 0.03 = (x - 0.02) + 0.01 \implies 0.03 = x - 0.01 \implies x = 0.04\text{ moles}$. The molarity of NaOH in $1\text{ L}$ is $0.04\text{ M}$. Therefore, correct answer is A.