Calculate pH solution:
0.1 M `CH_(3)COOH (K_a=1.8 xx 10^(-5))`

Answer

To calculate the pH of a 0.1 M acetic acid (CH₃COOH) solution with a given dissociation constant (Kₐ = 1.8 × 10⁻⁵), we can follow these steps:

### Step 1: Write the dissociation equation
Acetic acid dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \([CH_3COOH] = 0.1 \, M\)
- \([CH_3COO^-] = 0 \, M\)
- \([H^+] = 0 \, M\)

### Step 3: Define the change in concentration
Let \(x\) be the amount of acetic acid that dissociates. At equilibrium, the concentrations will be:
- \([CH_3COOH] = 0.1 - x\)
- \([CH_3COO^-] = x\)
- \([H^+] = x\)

### Step 4: Write the expression for the equilibrium constant (Kₐ)
The expression for the dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Substituting the equilibrium concentrations into the expression:
\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.1 - x} \]
This simplifies to:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \]

### Step 5: Assume \(x\) is small compared to 0.1 M
Since acetic acid is a weak acid, we can assume that \(x\) is small compared to 0.1 M. Thus, we can approximate:
\[ 0.1 - x \approx 0.1 \]
This simplifies our equation to:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]

### Step 6: Solve for \(x\)
Rearranging gives:
\[ x^2 = 1.8 \times 10^{-5} \times 0.1 \]
\[ x^2 = 1.8 \times 10^{-6} \]
Taking the square root:
\[ x = \sqrt{1.8 \times 10^{-6}} \]
\[ x \approx 1.34 \times 10^{-3} \]

### Step 7: Calculate the pH
Since \(x\) represents the concentration of \(H^+\):
\[ [H^+] = 1.34 \times 10^{-3} \, M \]
Now, we can calculate the pH:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(1.34 \times 10^{-3}) \]
Using logarithm properties:
\[ \text{pH} \approx 2.87 \]

### Final Answer
The pH of the 0.1 M acetic acid solution is approximately **2.87**.

---