Question

What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?

Answer 1

Answer:

1.75 L HNO₃

Explanation:

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

          = 17.5 g ÷ 39.997 g/mol

          = 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

             = 0.4375 moles ÷ 0.250 M

             = 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

Answer 2

1.75 L of 0.250 M HNO₃ is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water.

Let's consider the balanced neutralization equation.

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

First, we will convert 17.5 g of NaOH to moles using its molar mass (40.00 g/mol).

$17.5 g \times \frac{1mol}{40.00g} = 0.438 mol$

The molar ratio of NaOH to HNO₃ is 1:1. The moles of HNO₃ needed to react with 0.438 moles of NaOH are:

$0.438 mol NaOH \times \frac{1molHNO_3}{1molNaOH} = 0.438 mol HNO_3$

0.438 moles of HNO₃ are in a certain volume of a 0.250 M solution. The volume of the solution is:

$0.438 mol \times \frac{1L}{0.250mol} = 1.75 L$

1.75 L of 0.250 M HNO₃ is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water.

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