What volume (L)of 0.250 M HNO₃is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water? A)50.0 B)0.44 C)1.75 D)0.070 E)1.75 ×

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Quizplus · Accepted Answer

First, we need to find the number of moles of NaOH in the solution:

mass of NaOH = 17.5 g
molar mass of NaOH = 40.0 g/mol
number of moles of NaOH = (17.5 g )/(40.0 g/mol) = 0.4375 mol

Next, we can use the balanced chemical equation between HNO3 and NaOH to find the number of moles of HNO3 required to neutralize the NaOH:

HNO3 + NaOH → NaNO3 + H2O

The stoichiometry of this equation tells us that 1 mole of HNO3 reacts with 1 mole of NaOH. Therefore, we need 0.4375 mol of HNO3 to neutralize the NaOH.

Finally, we can use the definition of molarity to solve for volume:

Molarity = moles of solute / liters of solution

We rearrange this equation to solve for volume:

Liters of solution = moles of solute / molarity

Liters of solution = 0.4375 mol / 0.250 mol/L = 1.75 L

Therefore, the volume of 0.250 M HNO3 required to neutralize the solution is 1.75 L.

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