3.42 grams of Ba(OH)2 is placed in a flask, dissolved in water, and titrated with an HCI solution. It is found that 25.0mL of the HCI solution is required to just neutralize the Ba(OH)2. What is the molarity of the HCI solution?
Ba(OH)2 + 2HCl ==> 2H2O + BaCl2
mols Ba(OH)2 = M x L = ?
Convert mols Ba(OH)2 to mols HCl.
M HCl = mols HCl/L HCl
To find the molarity (M) of the HCl solution, we need to use the balanced equation and stoichiometry of the reaction between Ba(OH)2 and HCl.
The balanced equation for the reaction is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the equation, we can see that the molar ratio between Ba(OH)2 and HCl is 1:2.
First, we need to calculate the number of moles of Ba(OH)2 using its molar mass.
Molar mass of Ba(OH)2 = (1 × atomic mass of Ba) + (2 × atomic mass of O) + (2 × atomic mass of H)
= (1 × 137.33 g/mol) + (2 × 15.999 g/mol) + (2 × 1.008 g/mol)
= 137.33 g/mol + 31.998 g/mol + 2.016 g/mol
= 171.344 g/mol
Number of moles of Ba(OH)2 = mass / molar mass
= 3.42 g / 171.344 g/mol
≈ 0.02 mol
According to the stoichiometry, 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
So, the number of moles of HCl = 2 × number of moles of Ba(OH)2
= 2 × 0.02 mol
= 0.04 mol
Now, we can calculate the molarity of the HCl solution using the volume and number of moles of HCl.
Molarity (M) = number of moles / volume (in liters)
= 0.04 mol / 0.025 L
= 1.6 M
Therefore, the molarity of the HCl solution is 1.6 M.
To find the molarity of the HCl solution, we need to use the balanced chemical equation for the reaction between HCl and Ba(OH)2, as well as the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, we can see that one mole of Ba(OH)2 reacts with two moles of HCl.
First, we need to calculate the number of moles of Ba(OH)2 we have.
Given:
Mass of Ba(OH)2 = 3.42 grams
Molar mass of Ba(OH)2 = 171.34 g/mol (atomic masses: Ba = 137.33 g/mol, O = 16.00 g/mol, H = 1.01 g/mol)
Number of moles of Ba(OH)2 = mass / molar mass
= 3.42 g / 171.34 g/mol
≈ 0.02 mol Ba(OH)2
Now, we need to find the number of moles of HCl used in the titration.
Given:
Volume of HCl solution = 25.0 mL
(The volume needs to be converted to liters for consistent units in molarity)
Number of moles of HCl = (Volume of HCl solution in liters) x (Molarity of HCl)
We need to rearrange the equation to solve for the molarity of HCl.
Molarity of HCl = (Number of moles of HCl) / (Volume of HCl solution in liters)
To determine the number of moles of HCl, we can use the stoichiometry of the balanced equation. Since it is a 1:2 ratio between Ba(OH)2 and HCl, we can say:
Number of moles of HCl = 2 x Number of moles of Ba(OH)2
= 2 x 0.02 mol Ba(OH)2
= 0.04 mol HCl
Now we have all the information we need to calculate the molarity of the HCl solution:
Molarity of HCl = (0.04 mol HCl) / (25.0 mL HCl solution converted to liters)
= (0.04 mol) / (0.025 L)
= 1.6 M
Therefore, the molarity of the HCl solution is 1.6 M.