Calculate the molarity of a Ba(OH)2 solution if 1950 mL is completely titrated by 26.1 mL of 0.505 M HNO3
The Correct Answer and Explanation is:
To calculate the molarity of the Ba(OH)₂ solution, we will use the concept of neutralization reactions and the stoichiometry of the reaction between Ba(OH)₂ and HNO₃.
Step 1: Write the balanced chemical equation
The reaction between barium hydroxide (Ba(OH)₂) and nitric acid (HNO₃) is:
[
\text{Ba(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Ba(NO}_3)_2(aq) + 2\text{H}_2\text{O}(l)
]
This equation shows that one mole of Ba(OH)₂ reacts with two moles of HNO₃.
Step 2: Calculate the moles of HNO₃ used
We are given the volume and molarity of the HNO₃ solution. The formula to calculate moles is:
[
\text{moles of HNO}_3 = M \times V
]
Where:
- ( M ) = molarity of HNO₃ = 0.505 M
- ( V ) = volume of HNO₃ = 26.1 mL = 0.0261 L
[
\text{moles of HNO}_3 = 0.505 \, \text{mol/L} \times 0.0261 \, \text{L} = 0.0132 \, \text{mol}
]
Step 3: Use stoichiometry to calculate moles of Ba(OH)₂
From the balanced equation, we see that 2 moles of HNO₃ react with 1 mole of Ba(OH)₂. Therefore, the moles of Ba(OH)₂ can be calculated as:
[
\text{moles of Ba(OH)}_2 = \frac{\text{moles of HNO}_3}{2} = \frac{0.0132}{2} = 0.0066 \, \text{mol}
]
Step 4: Calculate the molarity of Ba(OH)₂
We are given the volume of Ba(OH)₂ solution as 1950 mL, which is 1.950 L. Molarity is defined as moles of solute per liter of solution:
[
M = \frac{\text{moles of Ba(OH)}_2}{\text{volume of Ba(OH)}_2 \, \text{in liters}}
]
[
M = \frac{0.0066 \, \text{mol}}{1.950 \, \text{L}} = 0.00338 \, \text{M}
]
Final Answer:
The molarity of the Ba(OH)₂ solution is approximately 0.00338 M.
Explanation:
In this problem, we used the stoichiometry of the neutralization reaction between Ba(OH)₂ and HNO₃ to find the molarity of the Ba(OH)₂ solution. By first determining the number of moles of HNO₃ used in the titration, we applied the molar ratio from the balanced equation to calculate the moles of Ba(OH)₂. Finally, we divided the moles of Ba(OH)₂ by the volume of the Ba(OH)₂ solution to find its molarity. This approach ensures we account for the chemical relationship between the acid and base and the volume of the solutions involved.