An aqueous solution of nitric acid is standardized by titration with a 0.121 M solution of calcium hydroxide.
If 24.5 mL of base are required to neutralize 15.4 mL of the acid, what is the molarity of the nitric acid solution?
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To find the molarity of the nitric acid solution, we can use the concept of stoichiometry in a titration.
In this titration, nitric acid reacts with calcium hydroxide in a 1:1 ratio, according to the balanced chemical equation:
2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
Given that 24.5 mL of the 0.121 M calcium hydroxide solution are required to neutralize 15.4 mL of the nitric acid solution, we can set up a ratio based on the balanced equation:
Moles of calcium hydroxide = Molarity × Volume = 0.121 M × 24.5 mL
Moles of nitric acid = Moles of calcium hydroxide (according to the balanced equation)
Since the ratio is 1:1, the moles of nitric acid will also be equal to 0.121 M × 24.5 mL.
Now, we can calculate the molarity of the nitric acid solution by dividing the moles of nitric acid by the volume of the nitric acid solution given:
Molarity of nitric acid = Moles of nitric acid / Volume of nitric acid solution
Molarity of nitric acid = (0.121 M × 24.5 mL) / 15.4 mL
Simplifying the calculation gives us the molarity of the nitric acid solution.
2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
mols Ca(OH)2 = M x L = ?
Convert mols Ca(OH)2 to mols HNO3 using the coefficients in the balanced equation.
M HNO3 = mols HNO3/L HNO3