Calculate the molarity of a Ba(OH)2 solution if 1.90 L is completely neutralized by 0.261 L of 0.505 M HNO3.
Well, I don't mean to be a "Ba-d" influence, but let's get down to business. First, we need to balance the equation:
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O.
Now, we can determine the moles of HNO3 using its concentration and volume:
moles of HNO3 = 0.261 L x 0.505 M = 0.131605 moles.
Since the stoichiometric ratio between HNO3 and Ba(OH)2 is 2:1, we can conclude that the moles of Ba(OH)2 in the solution are twice that of the HNO3:
moles of Ba(OH)2 = 2 x 0.131605 moles = 0.26321 moles.
Finally, we can determine the molarity (M) of the Ba(OH)2 solution by dividing the moles of Ba(OH)2 by the volume of the solution in liters:
Molarity = moles of Ba(OH)2 / volume of solution in liters
= 0.26321 moles / 1.90 L
= 0.1385 M.
So, the molarity of the Ba(OH)2 solution is approximately 0.1385 M.
To calculate the molarity of a Ba(OH)2 solution, we need to use the concept of stoichiometry and the volume and concentration of the acid used for neutralization.
1. Write the balanced chemical equation for the reaction between Ba(OH)2 and HNO3:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O
2. Determine the mole ratio between Ba(OH)2 and HNO3:
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3.
3. Calculate the number of moles of HNO3 used:
moles of HNO3 = volume of HNO3 used (in L) x concentration of HNO3 (in M)
moles of HNO3 = 0.261 L x 0.505 M
moles of HNO3 = 0.1318 mol
4. Use the mole ratio to determine the number of moles of Ba(OH)2:
Since the mole ratio is 1:2 (Ba(OH)2:HNO3), the number of moles of Ba(OH)2 is half the number of moles of HNO3.
moles of Ba(OH)2 = 0.1318 mol / 2
moles of Ba(OH)2 = 0.0659 mol
5. Calculate the molarity of the Ba(OH)2 solution:
Molarity = moles of solute / volume of solution (in L)
Molarity = 0.0659 mol / 1.90 L
Molarity = 0.0347 M
Therefore, the molarity of the Ba(OH)2 solution is 0.0347 M.
To calculate the molarity of a Ba(OH)2 solution, we need to use the concept of stoichiometry, which relates the number of moles of reactants and products in a chemical equation.
First, let's write the balanced chemical equation for the reaction between Ba(OH)2 and HNO3:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3 to form 1 mole of Ba(NO3)2.
Step 1: Calculate the number of moles of HNO3 used in the reaction.
Molarity of HNO3 (M1) = 0.505 M
Volume of HNO3 used (V1) = 0.261 L
Using the formula:
Number of moles = Molarity x Volume
Number of moles of HNO3 = 0.505 M x 0.261 L = 0.132005 moles (approximately 0.132 mol)
Step 2: Calculate the number of moles of Ba(OH)2 used in the reaction.
From the balanced equation, we know that 2 moles of HNO3 react with 1 mole of Ba(OH)2.
Therefore, the number of moles of Ba(OH)2 is equal to half the number of moles of HNO3 used.
Number of moles of Ba(OH)2 = 0.132005 moles / 2 = 0.066 moles
Step 3: Calculate the molarity of Ba(OH)2.
Molarity = Number of moles / Volume
Volume of Ba(OH)2 (V2) = 1.90 L
Molarity of Ba(OH)2 = 0.066 moles / 1.90 L = 0.0347 M (approximately)
Therefore, the molarity of the Ba(OH)2 solution is approximately 0.0347 M.
2HNO3 + Ba(OH)2 ==> Ba(NO3)2 + 2H2O
mols HNO3 = M x L = 0.505 x 0.261 = ?
From the coefficients in the balanced equation 2 mol HNO3 = 1 mol Ba(OH)2; therefore mols Ba(OH)2 = 1/2 mol HNO3
Then M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2 = ?
Post your work if you get stuck.