If the titration of a 25.0 mL sample of calcium hydroxide requires 34.45 mL of 0.100 M perchloric acid, what is the molarity of the base?
2 HClO4(aq) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + 2 H2O(l)
By definition, M Ca(OH)2 = mols/L solution.
mols HClO4 = M x L = ?
mols NaOH = 1/2 that (from the coefficients in the balanced equation).
Then M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2.
To find the molarity of the base (calcium hydroxide, Ca(OH)2), we can use the stoichiometry of the balanced equation.
First, let's write the balanced equation:
2 HClO4(aq) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + 2 H2O(l)
From the equation, we can see that the mole ratio between HClO4 and Ca(OH)2 is 2:1. In other words, it takes 2 moles of HClO4 to react with 1 mole of Ca(OH)2.
Given:
Volume of HClO4 = 34.45 mL = 0.03445 L
Molarity of HClO4 = 0.100 M
Using the equation Molarity (M) = moles of solute / volume of solution in liters, we can calculate the number of moles of HClO4:
moles of HClO4 = Molarity × volume
moles of HClO4 = 0.100 M × 0.03445 L
moles of HClO4 = 0.003445 moles
Since the mole ratio between HClO4 and Ca(OH)2 is 2:1, the number of moles of Ca(OH)2 is half that of HClO4:
moles of Ca(OH)2 = 0.003445 moles ÷ 2
moles of Ca(OH)2 = 0.001723 moles
Now, we can calculate the molarity of Ca(OH)2 using the equation Molarity = moles of solute / volume of solution in liters:
Molarity of Ca(OH)2 = moles of Ca(OH)2 / volume of Ca(OH)2 in liters
Volume of Ca(OH)2 = 25.0 mL = 0.0250 L
Molarity of Ca(OH)2 = 0.001723 moles / 0.0250 L
Molarity of Ca(OH)2 = 0.069 moles/L
Therefore, the molarity of the base (Ca(OH)2) is 0.069 M.
To find the molarity of the base (calcium hydroxide), we can use the balanced chemical equation and the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reaction:
2 HClO4(aq) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + 2 H2O(l)
From the balanced equation, we can see that the stoichiometric ratio between HClO4 and Ca(OH)2 is 2:1. This means that for every 2 moles of HClO4, we have 1 mole of Ca(OH)2.
In the given titration, 34.45 mL of 0.100 M HClO4 is required to react with the calcium hydroxide. To find the number of moles of HClO4 used, we can use the formula:
Moles of solute (HClO4) = Volume (in liters) × Molarity
Convert the volume of HClO4 to liters:
34.45 mL = 34.45 mL × (1 L / 1000 mL) = 0.03445 L
Now calculate the number of moles of HClO4:
Moles of HClO4 = 0.03445 L × 0.100 mol/L = 0.003445 mol
According to the stoichiometry of the balanced equation, the number of moles of Ca(OH)2 will be half of the moles of HClO4 used. Therefore, the number of moles of Ca(OH)2 can be calculated as follows:
Moles of Ca(OH)2 = 0.003445 mol / 2 = 0.001723 mol
Finally, to find the molarity of the base (calcium hydroxide), we divide the number of moles by the volume in liters:
Molarity of Ca(OH)2 = Moles of Ca(OH)2 / Volume (in liters)
The volume of the sample is given as 25.0 mL, so we convert it to liters:
25.0 mL = 25.0 mL × (1 L / 1000 mL) = 0.0250 L
Now calculate the molarity of Ca(OH)2:
Molarity of Ca(OH)2 = 0.001723 mol / 0.0250 L ≈ 0.069 M
Therefore, the molarity of the base (calcium hydroxide) is approximately 0.069 M.