1.75 L HNO₃
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
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).
[tex]17.5 g \times \frac{1mol}{40.00g} = 0.438 mol[/tex]
The molar ratio of NaOH to HNO₃ is 1:1. The moles of HNO₃ needed to react with 0.438 moles of NaOH are:
[tex]0.438 mol NaOH \times \frac{1molHNO_3}{1molNaOH} = 0.438 mol HNO_3[/tex]
0.438 moles of HNO₃ are in a certain volume of a 0.250 M solution. The volume of the solution is:
[tex]0.438 mol \times \frac{1L}{0.250mol} = 1.75 L[/tex]
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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