Calculate pH solution:
`10^(-8)M CH_(3)COOH (K_(a)=1.8 xx 10^(-5))`

Answer

To calculate the pH of a \(10^{-8} M\) solution of acetic acid (\(CH_3COOH\)), we need to consider both the dissociation of the acetic acid and the contribution of hydrogen ions from water. Here’s a step-by-step solution:

### Step 1: Understand the dissociation of acetic acid
Acetic acid is a weak acid and partially dissociates in water according to the equation:
\[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]
The dissociation constant (\(K_a\)) for acetic acid is given as \(1.8 \times 10^{-5}\).

### Step 2: Set up the equilibrium expression
Let \(x\) be the concentration of \(H^+\) ions produced from the dissociation of acetic acid. At equilibrium, the concentrations will be:
- \([H^+] = x\)
- \([CH_3COO^-] = x\)
- \([CH_3COOH] = 10^{-8} - x\)

The equilibrium expression for the dissociation of acetic acid can be written as:
\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{10^{-8} - x} \]

### Step 3: Substitute \(K_a\) and simplify
Substituting the value of \(K_a\):
\[ 1.8 \times 10^{-5} = \frac{x^2}{10^{-8} - x} \]
Since \(10^{-8}\) is very small compared to \(x\), we can approximate \(10^{-8} - x \approx 10^{-8}\):
\[ 1.8 \times 10^{-5} \approx \frac{x^2}{10^{-8}} \]
Thus,
\[ x^2 = 1.8 \times 10^{-5} \times 10^{-8} = 1.8 \times 10^{-13} \]

### Step 4: Solve for \(x\)
Taking the square root of both sides:
\[ x = \sqrt{1.8 \times 10^{-13}} \approx 1.34 \times 10^{-7} M \]

### Step 5: Calculate total \(H^+\) concentration
The total concentration of \(H^+\) ions in the solution will be the sum of the \(H^+\) ions from acetic acid and from water:
\[ [H^+]_{total} = x + 10^{-7} = 1.34 \times 10^{-7} + 1.0 \times 10^{-7} = 2.34 \times 10^{-7} M \]

### Step 6: Calculate pH
Now, we can calculate the pH using the formula:
\[ pH = -\log[H^+]_{total} = -\log(2.34 \times 10^{-7}) \approx 6.63 \]

### Final Answer
The pH of the \(10^{-8} M\) acetic acid solution is approximately **6.63**.