Calculate the molarity if a Ba(OH)2 solution if a 37.25ml portion of this solution is neutralized by 45.00mL of 1.500M HCl. You will need the balanced reaction
To calculate the molarity (M) of the Ba(OH)2 solution, we first need to write the balanced reaction between Ba(OH)2 and HCl.
The balanced reaction is as follows:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl. This means that the mole ratio between Ba(OH)2 and HCl is 1:2.
Given that 45.00 mL of 1.500 M HCl is neutralized by 37.25 mL of the Ba(OH)2 solution, we can calculate the number of moles of HCl used.
Moles of HCl = volume (in L) × molarity
Moles of HCl = 0.04500 L × 1.500 M
Moles of HCl = 0.0675 moles
Since the mole ratio between Ba(OH)2 and HCl is 1:2, the number of moles of Ba(OH)2 used is half of the moles of HCl.
Moles of Ba(OH)2 = 0.0675 moles ÷ 2
Moles of Ba(OH)2 = 0.03375 moles
Now, we can calculate the molarity of the Ba(OH)2 solution.
Molarity = moles of solute ÷ volume of solution (in L)
Molarity = 0.03375 moles ÷ 0.03725 L
Molarity ≈ 0.908 M
Therefore, the molarity of the Ba(OH)2 solution is approximately 0.908 M.
The balanced reaction between Ba(OH)2 and HCl is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
To calculate the molarity (M) of the Ba(OH)2 solution, we can use the concept of stoichiometry and the given volume and molarity of HCl.
Step 1: Identify the known values
Volume of Ba(OH)2 solution: 37.25 mL
Volume of HCl solution: 45.00 mL
Molarity of HCl: 1.500 M
Step 2: Convert the volume of HCl to moles
Using the formula: moles = volume (L) x molarity (M)
moles of HCl = 0.045 L x 1.500 M = 0.0675 moles
Step 3: Use the stoichiometry of the balanced equation to find moles of Ba(OH)2
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
So, moles of Ba(OH)2 = 0.0675 moles of HCl x (1 mole Ba(OH)2 / 2 moles HCl) = 0.03375 moles
Step 4: Convert moles to molarity
Molarity (M) is calculated by dividing moles (n) by volume (V) in liters.
Molarity of Ba(OH)2 = moles / volume (in liters) = 0.03375 moles / 0.03725 L = 0.906 M
Therefore, the molarity of the Ba(OH)2 solution is 0.906 M.