What volume of 0.0658 M calcium hydroxide is required to neutralize 45.10 mL of 0.0242 M nitric acid?
the famous titration equation:
Na*volumeAcid=Nb*volumeBase
where N is concentration in Normality. Normality equals Molarity times Heq+.
Nitric acid HNO3 is ONE H+
Ca(OH)2 is TWO H+ equivalents
1*.0658 Va=2*.0242*.45.10 ml
solve for Va
To find the volume of calcium hydroxide required to neutralize the nitric acid, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and nitric acid (HNO3).
The balanced chemical equation for the reaction between calcium hydroxide and nitric acid is:
2 HNO3 + Ca(OH)2 → Ca(NO3)2 + 2 H2O
From the balanced equation, we can see that 2 moles of nitric acid react with 1 mole of calcium hydroxide.
Step 1: Convert the given volume of nitric acid to moles
To do this, we can use the formula:
moles = concentration (in M) × volume (in L)
Given:
Concentration of nitric acid (HNO3) = 0.0242 M
Volume of nitric acid (HNO3) = 45.10 mL = 45.10/1000 L = 0.04510 L
moles of HNO3 = 0.0242 M × 0.04510 L = 0.00109142 moles
Step 2: Use the mole ratio from the balanced equation to calculate moles of calcium hydroxide
From the balanced equation, we know that 2 moles of nitric acid react with 1 mole of calcium hydroxide.
So, moles of Ca(OH)2 = 0.00109142/2 = 0.00054571 moles
Step 3: Convert moles of calcium hydroxide to volume
To do this, we need to use the concentration of calcium hydroxide (Ca(OH)2).
Given:
Concentration of calcium hydroxide (Ca(OH)2) = 0.0658 M
volume = moles / concentration
volume = 0.00054571 moles / 0.0658 M ≈ 0.00828 L = 8.28 mL
Therefore, the volume of 0.0658 M calcium hydroxide required to neutralize 45.10 mL of 0.0242 M nitric acid is approximately 8.28 mL.