pH of a 0.01 M solution `(K_(a)=6.6xx10^(-4))`

Options

A. 7.6
B. 8
C. 2.6
D. 5

Explanation

To find the pH of a 0.01 M solution with a given \( K_a = 6.6 \times 10^{-4} \), we can follow these steps:

### Step 1: Write the expression for the dissociation of the weak acid.
Assuming the weak acid is HA, it dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the equilibrium expression.
Let the initial concentration of HA be \( C = 0.01 \, \text{M} \). At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- Concentration of \( H^+ \) at equilibrium = \( C \alpha \)
- Concentration of \( A^- \) at equilibrium = \( C \alpha \)
- Concentration of undissociated HA = \( C(1 - \alpha) \)

### Step 3: Write the expression for \( K_a \).
The expression for the acid dissociation constant \( K_a \) is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} \]
\[ K_a = \frac{C^2 \alpha^2}{C(1 - \alpha)} \]
\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]

### Step 4: Assume \( \alpha \) is small.
Since \( K_a \) is small, we can assume \( \alpha \) is small compared to 1, thus \( 1 - \alpha \approx 1 \):
\[ K_a \approx C \alpha^2 \]

### Step 5: Solve for \( \alpha \).
Rearranging gives:
\[ \alpha^2 = \frac{K_a}{C} \]
Substituting the values:
\[ \alpha^2 = \frac{6.6 \times 10^{-4}}{0.01} = 6.6 \times 10^{-2} \]
Taking the square root:
\[ \alpha = \sqrt{6.6 \times 10^{-2}} \approx 0.257 \]

### Step 6: Calculate the concentration of \( H^+ \).
Using the value of \( \alpha \):
\[ [H^+] = C \alpha = 0.01 \times 0.257 = 2.57 \times 10^{-3} \, \text{M} \]

### Step 7: Calculate the pH.
Using the formula for pH:
\[ pH = -\log[H^+] = -\log(2.57 \times 10^{-3}) \]
Using logarithmic properties:
\[ pH = -(\log(2.57) + \log(10^{-3})) \]
\[ pH = 3 - \log(2.57) \]
Calculating \( \log(2.57) \) gives approximately \( 0.411 \):
\[ pH \approx 3 - 0.411 = 2.589 \]
Rounding gives:
\[ pH \approx 2.6 \]

### Final Answer:
The pH of the solution is approximately **2.6**.

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