The volume of 1.5 M solution required to neutralize exactly 90 mL of a 0.5 M solution is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

According to the Law of Equivalence: $N_1V_1 = N_2V_2$. For $H_3PO_4$, n-factor = 3, so $N_1 = 1.5 \times 3 = 4.5 N$. For $Ba(OH)_2$, n-factor = 2, so $N_2 = 0.5 \times 2 = 1.0 N$. $4.5 \times V_1 = 1.0 \times 90 \Rightarrow 4.5V_1 = 90 \Rightarrow V_1 = 20$ mL.