What volume of a 0.442 M NaOH solution is needed to neutralize 65.0 mL of a 0.296 M solution of HNO₃? A) 8.71 mL B) 21.8 mL C) 43.5 mL D) 87.1 mL E) 174 mL

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To solve this problem, we use the concept of molarity and the stoichiometry of the neutralization reaction. The reaction between NaOH and HNO₃ (assuming it behaves similarly to a typical acid HX) is:$$ \text{NaOH} + \text{HX} \rightarrow \text{NaX} + \text{H}_2\text{O} $$This reaction shows that one mole of NaOH reacts with one mole of HX to neutralize it. Therefore, the moles of NaOH needed equals the moles of HX present.First, calculate the moles of HX:$$ \text{Moles of HX} = 0.296 \, \text{M} \times 0.065 \, \text{L} = 0.01924 \, \text{moles} $$Since the stoichiometry of the reaction is 1:1, the moles of NaOH needed to neutralize HX are also 0.01924 moles.Next, calculate the volume of NaOH solution needed:$$ \text{Volume of NaOH} = \frac{\text{Moles of NaOH}}{\text{Molarity of NaOH}} = \frac{0.01924}{0.442} \, \text{L} = 0.0435 \, \text{L} $$Converting to mL (since 1 L = 1000 mL):$$ 0.0435 \, \text{L} \times 1000 = 43.5 \, \text{mL} $$Therefore, 43.5 mL of the 0.442 M NaOH solution is needed to neutralize 65.0 mL of the 0.296 M solution of HX.

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