The volume of required to neutralize mL of completely is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

For complete neutralization: $n_1 M_1 V_1 (\text{acid}) = n_2 M_2 V_2 (\text{base})$. <br>$H_3PO_4$ ($n=3$), $Ca(OH)_2$ ($n=2$). <br>$3 \times 0.1 \times 10 = 2 \times 0.1 \times V \Rightarrow 3 = 0.2 V \Rightarrow V = 15$ mL.