For a 0.1 M aqueous solution of a weak acis, HA `(K_(a)=10^(-9))`, the pH is approximately equal to

Options

A. 9
B. 3
C. 11
D. 10

Answer

To find the pH of a 0.1 M aqueous solution of a weak acid HA with a dissociation constant \( K_a = 10^{-9} \), we can follow these steps:

### Step 1: Write the dissociation equation
The weak acid HA dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the equilibrium expression
The equilibrium expression for the dissociation of the acid is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

### Step 3: Define the initial concentrations
Let:
- The initial concentration of HA = \( C = 0.1 \, M \)
- At equilibrium, let the concentration of \( H^+ \) and \( A^- \) be \( x \).

Thus, at equilibrium:
- \([H^+] = x\)
- \([A^-] = x\)
- \([HA] = C - x \approx C\) (since \( x \) will be very small compared to \( C \))

### Step 4: Substitute into the \( K_a \) expression
Substituting the equilibrium concentrations into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{C} = \frac{x^2}{C} \]
\[ 10^{-9} = \frac{x^2}{0.1} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = 10^{-9} \cdot 0.1 = 10^{-10} \]
\[ x = \sqrt{10^{-10}} = 10^{-5} \]

### Step 6: Find the concentration of \( H^+ \)
Thus, the concentration of \( H^+ \) ions is:
\[ [H^+] = x = 10^{-5} \, M \]

### Step 7: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ \text{pH} = -\log(10^{-5}) = 5 \]

### Conclusion
Therefore, the pH of the 0.1 M aqueous solution of the weak acid HA is approximately 5.

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