What is the molarity of a calcium hydroxide solution if 29.52 mL of 0.4 M nitric acid is needed to neutralize 0.046 L
Ca(OH)2 + 2HNO3 ==> 2H2O + Ca(NO3)2
mols HNO3 = M x L = 0.4 x 0.02952 = 0.01181
Convert mole HNO3 to moles Ca(OH)2. 0.01181/2 = 0.005905
M Ca(OH)2 = moles/L = 0.005592 moles/0.046 L = 0.1283 M
It is 0.2 M?
To determine the molarity of the calcium hydroxide solution, we can use the equation for the neutralization reaction between calcium hydroxide (Ca(OH)2) and nitric acid (HNO3):
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
In this reaction, 1 mole of calcium hydroxide (Ca(OH)2) reacts with 2 moles of nitric acid (HNO3) to produce 1 mole of calcium nitrate (Ca(NO3)2) and 2 moles of water (H2O).
First, let's determine the number of moles of nitric acid used in the reaction. We can use the formula:
moles = volume (L) × molarity
moles of nitric acid = 29.52 mL × 0.4 M / 1000 mL/L
moles of nitric acid = 0.011808 mol
Since the stoichiometry of the reaction tells us that it takes 2 moles of nitric acid to neutralize 1 mole of calcium hydroxide, the number of moles of calcium hydroxide can be calculated as follows:
moles of calcium hydroxide = 0.011808 mol / 2
moles of calcium hydroxide = 0.005904 mol
Now, let's determine the volume of the calcium hydroxide solution used in the reaction. We are given that the volume is 0.046 L.
Finally, we can calculate the molarity of the calcium hydroxide solution using the formula:
molarity = moles / volume
molarity of calcium hydroxide solution = 0.005904 mol / 0.046 L
molarity of calcium hydroxide solution ≈ 0.128 M
Therefore, the molarity of the calcium hydroxide solution is approximately 0.128 M.