The hydrogen ion concentration of a 0.006 M benzoic acid solution is `(K_(a) = 6 xx 10^(-5))`

Options

A. `0.6 xx 10^(-4)`
B. `6 xx 10^(-4)`
C. `6 xx 10^(-5)`
D. `3.6 xx 10^(-4)`

Answer

To find the hydrogen ion concentration of a 0.006 M benzoic acid solution, we can follow these steps:

### Step 1: Write the dissociation equation
Benzoic acid (C6H5COOH) dissociates in water to form benzoate ions (C6H5COO-) and hydrogen ions (H+):
\[ \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the expression for the acid dissociation constant (Ka)
The expression for the acid dissociation constant (Ka) is given by:
\[ K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \]

### Step 3: Define the initial concentrations and changes
Let the initial concentration of benzoic acid be \( C_0 = 0.006 \) M. At equilibrium, let \( x \) be the concentration of H+ ions produced. Therefore, at equilibrium:
- \([\text{C}_6\text{H}_5\text{COOH}] = 0.006 - x\)
- \([\text{C}_6\text{H}_5\text{COO}^-] = x\)
- \([\text{H}^+] = x\)

### Step 4: Substitute into the Ka expression
Substituting the equilibrium concentrations into the Ka expression gives:
\[ K_a = \frac{x \cdot x}{0.006 - x} = \frac{x^2}{0.006 - x} \]

### Step 5: Simplify the equation
Since \( x \) is expected to be much smaller than 0.006, we can approximate:
\[ 0.006 - x \approx 0.006 \]
Thus, the equation simplifies to:
\[ K_a = \frac{x^2}{0.006} \]

### Step 6: Substitute the value of Ka
Given that \( K_a = 6 \times 10^{-5} \):
\[ 6 \times 10^{-5} = \frac{x^2}{0.006} \]

### Step 7: Solve for x
Rearranging the equation gives:
\[ x^2 = 6 \times 10^{-5} \times 0.006 \]
\[ x^2 = 36 \times 10^{-8} \]

### Step 8: Calculate x
Taking the square root of both sides:
\[ x = \sqrt{36 \times 10^{-8}} = 6 \times 10^{-4} \]

### Conclusion
The hydrogen ion concentration \([H^+]\) in the 0.006 M benzoic acid solution is:
\[ [H^+] = 6 \times 10^{-4} \, \text{M} \]