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What is the molarity of a solution of nitric acid if 0.167 g of barium hydroxide is required to neutralize 16.99 mL of nitric acid?

Question

What is the molarity of a solution of nitric acid if 0.167 g of barium hydroxide is required to neutralize 16.99 mL of nitric acid? 

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Step 1

The balanced chemical equation between barium hydroxide and nitric acid is shown as,

Ba (OH) 2HNO,
Ba (NO,), +2H,0
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Ba (OH) 2HNO, Ba (NO,), +2H,0

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Step 2

To calculate the molarity, it is required to calculate the moles of nitric acid, which can be calculated as,

Mass
Moles of Barium hydroxide
Molar mass
Moles of Barium hydroxide =
9.74 x 10-4 moles
171.34
1 mole of Barium hydroxide requires 2 moles of nitric acid
Therefore, 9.74 x 104 moles of barium hydroxide will require 2 x 9.74 x 104
moles of nitric acid
Moles of nitric acid = 2 x 9.74 x 10-4 1.94 x 103 moles.
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Mass Moles of Barium hydroxide Molar mass Moles of Barium hydroxide = 9.74 x 10-4 moles 171.34 1 mole of Barium hydroxide requires 2 moles of nitric acid Therefore, 9.74 x 104 moles of barium hydroxide will require 2 x 9.74 x 104 moles of nitric acid Moles of nitric acid = 2 x 9.74 x 10-4 1.94 x 103 moles.

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Step 3

Now, calculating the molarit...

moles of solute
MolarityVolume of solution (1)
1.94 x 10-3
Molarity 16.99 x 10-3 0.114 M HNO3
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moles of solute MolarityVolume of solution (1) 1.94 x 10-3 Molarity 16.99 x 10-3 0.114 M HNO3

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