If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. Show all of the work used to solve this problem.
Unbalanced equation: Ba(OH)2 + HNO3 -> Ba(NO3)2 + H2O
To solve for the molarity of the nitric acid solution, we can use the following formula:
Molarity = (moles of solute) / (volume of solution in liters)
1. Start by balancing the equation:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O
2. Convert the given volumes to liters:
Volume of barium hydroxide solution = 23.4 mL = 23.4 / 1000 = 0.0234 L
Volume of nitric acid solution = 42.5 mL = 42.5 / 1000 = 0.0425 L
3. Use the stoichiometry of the balanced equation to find the number of moles of nitric acid:
From the balanced equation, the stoichiometric ratio between barium hydroxide and nitric acid is 1:2.
Therefore, the moles of nitric acid = (moles of barium hydroxide) x 2
Moles of barium hydroxide = Molarity x Volume (in liters)
Moles of barium hydroxide = 0.65 mol/L x 0.0234 L
Moles of nitric acid = 0.65 mol/L x 0.0234 L x 2
4. Calculate the molarity of the nitric acid solution:
Molarity = (moles of nitric acid) / (volume of nitric acid solution in liters)
Molarity = (0.65 mol/L x 0.0234 L x 2) / 0.0425 L
Molarity = 0.7317647 M
So, the molarity of the nitric acid solution is approximately 0.7318 M.
To solve this problem, we need to use the equation and the given information to determine the molarity of the nitric acid solution.
Let's start by calculating the number of moles of barium hydroxide (Ba(OH)2) that were used in the neutralization reaction.
Step 1: Calculate the number of moles of Ba(OH)2
Given:
- Volume of barium hydroxide solution (V1) = 23.4 mL
- Molarity of barium hydroxide solution (M1) = 0.65 M
Using the definition of molarity, we can convert the volume to liters:
V1 = 23.4 mL * (1 L / 1000 mL) = 0.0234 L
The number of moles (n1) can be calculated using the equation:
n1 = M1 * V1
Substituting the values, we get:
n1 = 0.65 M * 0.0234 L = 0.01521 moles
Therefore, 23.4 milliliters of 0.65 molar barium hydroxide solution contains 0.01521 moles of Ba(OH)2.
Step 2: Determine the stoichiometric ratio between Ba(OH)2 and HNO3
From the balanced equation:
1 mole of Ba(OH)2 reacts with 2 moles of HNO3
Step 3: Calculate the number of moles of HNO3
Since the stoichiometric ratio is 1:2 between Ba(OH)2 and HNO3, the number of moles of HNO3 (n2) can be calculated as follows:
n2 = 2 * n1
Substituting the value of n1, we get:
n2 = 2 * 0.01521 moles = 0.03042 moles
Therefore, 42.5 milliliters of nitric acid contains 0.03042 moles of HNO3.
Step 4: Calculate the molarity of nitric acid solution (M2)
The molarity (M2) can be calculated using the equation:
M2 = n2 / V2
Given:
- Volume of nitric acid solution (V2) = 42.5 mL = 0.0425 L
Substituting the values, we get:
M2 = 0.03042 moles / 0.0425 L = 0.7162 M
Therefore, the molarity of the nitric acid solution is approximately 0.7162 M.