The molarity of an aqueous solution of barium hydroxide is determined by titration against a 0.335 M nitric acid solution.
If 32.8 mL of the base are required to neutralize 21.5 mL of nitric acid, what is the molarity of the barium hydroxide solution?
Ba(OH)2 + 2HNO3 ==> Ba(NO3)2 + 2H2O
mols HNO3 = M x L = ?
mols Ba(OH)2 = 1/2 that from the coefficients in the balanced equation.
Then M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2. YOu know mols and L, solve for M.
To find the molarity of the barium hydroxide solution, we can use the concept of stoichiometry and the known volume and concentration of the nitric acid solution used in the titration.
In this titration, the reaction is:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O
From the balanced equation, we can see that 1 mole of barium hydroxide (Ba(OH)2) reacts with 2 moles of nitric acid (HNO3) to form 1 mole of barium nitrate (Ba(NO3)2) and 2 moles of water (H2O).
First, let's calculate the number of moles of nitric acid used in the reaction:
Moles of HNO3 = Volume (in L) x Concentration (in mol/L)
= 21.5 mL x (1 L / 1000 mL) x 0.335 mol/L
= 0.007209 mol
From the balanced equation, we can see that 1 mole of barium hydroxide reacts with 2 moles of nitric acid. Therefore, the number of moles of barium hydroxide is twice the number of moles of nitric acid used:
Moles of Ba(OH)2 = 2 x Moles of HNO3
= 2 x 0.007209 mol
= 0.01442 mol
Now, we can calculate the molarity of the barium hydroxide solution:
Molarity of Ba(OH)2 solution (in mol/L) = Moles of Ba(OH)2 / Volume (in L)
= 0.01442 mol / (32.8 mL x (1 L / 1000 mL))
≈ 0.439 mol/L
Therefore, the molarity of the barium hydroxide solution is approximately 0.439 mol/L.