What volume of 0.1 M is needed to exactly neutralize 40 mL of 0.2 M NaOH? — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

$M_1V_1 \times n_1 = M_2V_2 \times n_2 \Rightarrow 0.1 \times V \times 2 = 0.2 \times 40 \times 1 \Rightarrow 0.2V = 8 \Rightarrow V = 40$. Wait, $0.2 \times 40 / 0.2 = 40$. Let's re-calculate. $H_2SO_4$ has $n=2$. $0.1 \times V \times 2 = 0.2 \times 40 \Rightarrow 0.2V = 8 \Rightarrow V=40 \text{ mL}$. Correct Answer: A. [8]