The `P^(H)` of 0.001 M C H 3 C O O H is

Options

A. 3
B. 11
C. Between 3 and 7
D. 7

Answer

To calculate the pH of a 0.001 M solution of acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Understand the dissociation of acetic acid
Acetic acid is a weak acid that partially dissociates in solution according to the following equilibrium reaction:

\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the equilibrium expression
For a weak acid like acetic acid, we can use the acid dissociation constant (Ka) to express the equilibrium concentration of the ions. The dissociation constant for acetic acid (Ka) is approximately \(1.8 \times 10^{-5}\).

### Step 3: Write the expression for Ka
The expression for Ka is given by:

\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

### Step 4: Assume initial concentrations
Let the initial concentration of acetic acid be 0.001 M. At equilibrium, let x be the concentration of H⁺ ions produced:

- \([\text{CH}_3\text{COOH}] = 0.001 - x\)
- \([\text{H}^+] = x\)
- \([\text{CH}_3\text{COO}^-] = x\)

### Step 5: Substitute into the Ka expression
Substituting these values into the Ka expression gives:

\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.001 - x} \]

Assuming \(x\) is small compared to 0.001, we can simplify this to:

\[ 1.8 \times 10^{-5} = \frac{x^2}{0.001} \]

### Step 6: Solve for x
Rearranging the equation gives:

\[ x^2 = 1.8 \times 10^{-5} \times 0.001 \]

\[ x^2 = 1.8 \times 10^{-8} \]

Taking the square root of both sides:

\[ x = \sqrt{1.8 \times 10^{-8}} \]

\[ x \approx 1.34 \times 10^{-4} \, \text{M} \]

### Step 7: Calculate the pH
Now that we have the concentration of H⁺ ions, we can calculate the pH:

\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the value of x:

\[ \text{pH} = -\log(1.34 \times 10^{-4}) \]

Calculating this gives:

\[ \text{pH} \approx 3.87 \]

### Final Answer
The pH of a 0.001 M solution of acetic acid is approximately 3.87.

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