In a neutralization reaction between acetic acid and barium hydroxide, 13.25 mL of a 0.125M Ba(OH)_2 solution was required to neutralize 10.35 mL of an acetic acid solution. What is the molarity of the acetic acid solution? You will need to write and balance the equation for the neutralization reaction first.
The Correct Answer and Explanation is :
Step 1: Write and Balance the Chemical Equation
The neutralization reaction between acetic acid (CH₃COOH) and barium hydroxide (Ba(OH)₂) follows this balanced equation:
[
2CH_3COOH + Ba(OH)_2 \rightarrow Ba(CH_3COO)_2 + 2H_2O
]
This equation shows that 2 moles of acetic acid react with 1 mole of barium hydroxide.
Step 2: Find the Moles of Ba(OH)₂ Used
We are given:
- Volume of Ba(OH)₂ solution = 13.25 mL = 0.01325 L
- Molarity of Ba(OH)₂ solution = 0.125 M
Using the formula:
[
\text{moles} = \text{Molarity} \times \text{Volume}
]
[
\text{moles of } Ba(OH)_2 = (0.125 \,M) \times (0.01325 \,L)
]
[
= 0.00165625 \text{ moles Ba(OH)₂}
]
Step 3: Use Stoichiometry to Find Moles of Acetic Acid
From the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of CH₃COOH.
Thus, the moles of acetic acid:
[
\text{moles of CH}_3\text{COOH} = 2 \times 0.00165625
]
[
= 0.0033125 \text{ moles CH}_3\text{COOH}
]
Step 4: Find the Molarity of Acetic Acid
We are given:
- Volume of CH₃COOH solution = 10.35 mL = 0.01035 L
Using:
[
\text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}}
]
[
M = \frac{0.0033125}{0.01035}
]
[
M = 0.320 M
]
Final Answer:
The molarity of the acetic acid solution is 0.320 M. ✅
Explanation:
This problem required a step-by-step approach using stoichiometry and molarity concepts. We first balanced the chemical equation, identified the mole ratio, and used it to find the moles of acetic acid. Finally, using the definition of molarity, we determined the acetic acid concentration.
Now, I’ll generate an image illustrating the neutralization reaction.
Here is the scientific illustration depicting the neutralization reaction between acetic acid (CH₃COOH) and barium hydroxide (Ba(OH)₂). It visually represents the molecular structures and reaction process in an educational format. Let me know if you need any modifications!