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)1.25 B)0.11 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, using its molar mass and given mass:

molar mass NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol 
moles NaOH = 17.5 g / 40.00 g/mol = 0.4375 mol

Next, we can use the balanced chemical equation for the neutralization reaction between NaOH and HNO3:

NaOH + HNO3 -> NaNO3 + H2O

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

Finally, we can use the definition of molarity to relate the number of moles of HNO3 to the volume and concentration of the HNO3 solution:

Molarity = moles / volume

We can rearrange this equation to solve for volume:

volume = moles / Molarity

volume = 0.4375 mol / 0.250 mol/L = 1.75 L

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

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