When 100 mL of 0.4 M `CH_3COOH` are mixed with 100 mL of 0.2 NaOH, the `[H_3O^+]` in the solution is approximately : `[K_a(CH_3COOH)=1.8xx10^(-5)]`

Options

A. `1.8xx10^(-6) M`
B. `1.8xx10^(-5) M`
C. `9xx10^(-6) M`
D. `9xx10^(-5) M`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of CH₃COOH and NaOH
- **Moles of CH₃COOH**:
\[ \text{Volume} = 100 \, \text{mL} = 0.1 \, \text{L} \]
\[ \text{Molarity} = 0.4 \, \text{M} \]
\[ \text{Moles of CH₃COOH} = \text{Molarity} \times \text{Volume} = 0.4 \, \text{mol/L} \times 0.1 \, \text{L} = 0.04 \, \text{mol} \]

- **Moles of NaOH**:
\[ \text{Volume} = 100 \, \text{mL} = 0.1 \, \text{L} \]
\[ \text{Molarity} = 0.2 \, \text{M} \]
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

### Step 2: Determine the reaction between CH₃COOH and NaOH
The reaction is:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining moles after the reaction
- **Initial moles of CH₃COOH**: 0.04 mol
- **Initial moles of NaOH**: 0.02 mol

Since NaOH is the limiting reagent, it will react completely:
- **Moles of CH₃COOH remaining**:
\[ 0.04 \, \text{mol} - 0.02 \, \text{mol} = 0.02 \, \text{mol} \]
- **Moles of NaOH remaining**: 0 (it is completely consumed)

### Step 4: Calculate the concentration of CH₃COOH in the new solution
The total volume after mixing is:
\[ 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]
- **Concentration of CH₃COOH**:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.02 \, \text{mol}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the concentration of H₃O⁺ using the dissociation of CH₃COOH
The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]

Using the \( K_a \) expression:
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Assuming \( x \) is the concentration of \( \text{H}^+ \) produced,
\[ K_a = 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \]
Assuming \( x \) is small compared to 0.1, we can simplify:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]
\[ x^2 = 1.8 \times 10^{-6} \]
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \, \text{M} \]

Thus, the concentration of \( [H_3O^+] \) in the solution is approximately \( 1.34 \times 10^{-3} \, \text{M} \).

### Final Answer:
The concentration of \( [H_3O^+] \) in the solution is approximately \( 1.34 \times 10^{-3} \, \text{M} \).

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