What volume (in mL)of 0.0550 M calcium hydroxide is required to neutralize 38.00 mL of 0.0700 M nitric acid?
2HNO3 + Ca(OH)2 ==>Ca(NO3)2 + 2H2O
Moles HNO3 initially = M x L
Moles Ca(OH)2 required is 2 x moles HNO3 (because o the equation 1 mol Ca(OH)2 = 2 moles HNO3).
Then moles Ca(OH)2 = L x M
You know moles and M, calculate L, then multiply by 1000 to convert to mL.
To calculate the volume of calcium hydroxide required to neutralize the given amount of nitric acid, we need to use the concept of stoichiometry and the reaction equation between calcium hydroxide and nitric acid.
The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and nitric acid (HNO3) can be written as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
From the balanced equation, we can see that one mole of calcium hydroxide reacts with two moles of nitric acid. This means that the ratio of calcium hydroxide to nitric acid in the reaction is 1:2.
Given:
- Concentration of the nitric acid (HNO3): 0.0700 M
- Volume of nitric acid (HNO3): 38.00 mL
- Concentration of calcium hydroxide (Ca(OH)2): 0.0550 M
To solve this problem, we can use the equation:
(Molarity of acid) × (Volume of acid) = (Molarity of base) × (Volume of base)
Let's substitute the given values into the equation:
(0.0700 M) × (38.00 mL) = (0.0550 M) × (Volume of calcium hydroxide)
Now, let's solve for the volume of calcium hydroxide:
Volume of calcium hydroxide = (0.0700 M × 38.00 mL) / 0.0550 M
= (2.66 mol/L × 0.03800 L) / 0.0550 mol/L
= 1.84 mL
Therefore, the volume of 0.0550 M calcium hydroxide required to neutralize 38.00 mL of 0.0700 M nitric acid is approximately 1.84 mL.