The volume of solution required to exactly neutralize 90 mL of a solution is: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
The volume of $1.5 M$ $H_3PO_4$ solution required to exactly neutralize 90 mL of a $0.5 M$ $Ba(OH)_2$ solution is:
Answer: C
💡 Solution & Explanation
According to the Law of Equivalence: $N_1V_1 = N_2V_2$ . For $H_3PO_4$ (assuming full neutralization), 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.
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