What is the `pH` of a `1 M CH_(3)COOH` a solution `K_(a)` of acetic acid `=1.8xx10^(-5)`?
`K=10^(-14)mol^(2) litre^(-2)`

Options

A. `9.4`
B. `4.8`
C. `3.6`
D. `2.4`

Explanation

To find the pH of a 1 M acetic acid solution given the dissociation constant \( K_a = 1.8 \times 10^{-5} \) and the ion product of water \( K_w = 10^{-14} \), we can follow these steps:

### Step 1: Write the dissociation equation for acetic acid
Acetic acid (\( CH_3COOH \)) dissociates in water as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

### Step 2: Set up the equilibrium expression
The equilibrium expression for the dissociation of acetic acid is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Let \( x \) be the concentration of \( H^+ \) ions that dissociate from acetic acid. Initially, the concentration of acetic acid is 1 M, and at equilibrium, it will be:
\[ [CH_3COOH] = 1 - x, \quad [CH_3COO^-] = x, \quad [H^+] = x \]
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{1 - x} = \frac{x^2}{1 - x} \]

### Step 3: Assume \( x \) is small
Since \( K_a \) is small, we can assume that \( x \) is much smaller than 1, which allows us to simplify the expression to:
\[ K_a \approx \frac{x^2}{1} = x^2 \]

### Step 4: Solve for \( x \)
Now substituting the value of \( K_a \):
\[ 1.8 \times 10^{-5} = x^2 \]
Taking the square root of both sides:
\[ x = \sqrt{1.8 \times 10^{-5}} \approx 4.24 \times 10^{-3} \, \text{M} \]

### Step 5: Calculate the pOH
Since \( x \) represents the concentration of \( H^+ \), we can find the pH:
\[ pH = -\log[H^+] = -\log(4.24 \times 10^{-3}) \approx 2.37 \]

### Step 6: Calculate the pOH and then pH
Next, we can find the pOH using the relationship:
\[ pOH = 14 - pH \]
Thus:
\[ pOH = 14 - 2.37 = 11.63 \]

### Step 7: Final pH calculation
Now, we can find the final pH:
\[ pH = 14 - pOH = 14 - 11.63 = 2.37 \]

### Conclusion
The pH of a 1 M acetic acid solution is approximately **2.37**.