What volume of 0.1060 M NaOH is needed to neutralize a 50.00 mL sample of 0.09500 M HNO₃? A) 55.79 mL B) 55.19 mL C) 50.00 mL D) 44.81 mL E) 5.19 mL

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To find the volume of NaOH needed to neutralize the given volume of HNO3, we use the concept of molarity and the stoichiometry of the reaction. The reaction between NaOH and HNO3 is a 1:1 stoichiometry, meaning 1 mole of NaOH neutralizes 1 mole of HNO3.First, calculate the moles of HNO3 in the 50.00 mL sample:Moles of HNO3 = Molarity of HNO3 * Volume of HNO3 (in liters) = 0.09500 M * 50.00 mL * (1 L / 1000 mL) = 0.00475 molesSince the reaction is 1:1, the moles of NaOH needed will also be 0.00475.Next, calculate the volume of 0.1060 M NaOH needed to provide 0.00475 moles:Volume of NaOH = Moles of NaOH / Molarity of NaOH = 0.00475 moles / 0.1060 M = 0.04481 liters, or 44.81 mL.

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