20 ml 0.1M `CH_(3),COOH` has `P^(H)` value 3. It is titrated with 0.1M NaOH
What is `K_(a) `of `CH_(3)COOH`

Options

A. `10^(-5)`
B. `10^(-6)`
C. `10^(-7)`
D. `10^(-8)`

Answer

To find the \( K_a \) of acetic acid (\( CH_3COOH \)), we can follow these steps:

### Step 1: Calculate the concentration of \( H^+ \) ions
Given the pH of the solution is 3, we can calculate the concentration of hydrogen ions (\( [H^+] \)) using the formula:

\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Write the expression for \( K_a \)
The dissociation of acetic acid in water can be represented as follows:

\[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]

For this equilibrium, the expression for the acid dissociation constant \( K_a \) is given by:

\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]

### Step 3: Determine the concentrations at equilibrium
Initially, the concentration of acetic acid (\( CH_3COOH \)) is 0.1 M, and the volume is 20 mL. Thus, the moles of acetic acid are:

\[ \text{Moles of } CH_3COOH = 0.1 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{moles} \]

At equilibrium, let \( x = [H^+] = 10^{-3} \, \text{M} \). Therefore, the concentration of \( CH_3COO^- \) will also be \( 10^{-3} \, \text{M} \) since it is produced in a 1:1 ratio. The concentration of \( CH_3COOH \) will be:

\[ [CH_3COOH] = 0.1 - x \approx 0.1 \, \text{M} \, (\text{since } x \text{ is very small compared to } 0.1) \]

### Step 4: Substitute values into the \( K_a \) expression
Now we can substitute the values into the \( K_a \) expression:

\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{(10^{-3})(10^{-3})}{0.1} \]

### Step 5: Calculate \( K_a \)
Calculating \( K_a \):

\[ K_a = \frac{10^{-6}}{0.1} = 10^{-5} \]

### Conclusion
Thus, the \( K_a \) of acetic acid is:

\[ K_a = 1.0 \times 10^{-5} \]

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