Ionisation constant of `CH_(3)COOH` is `1.7xx10^(-5)` and concentration of `H^(+)ions` is `3.4xx10^(-4)`. Then, find out initial concentration of `CH_(3)COOH` molecules.

Options

A. `3.4 xx10^(-4)`
B. `3.4 xx 10^(-3)`
C. `6.8xx10^(-4)`
D. `6.8 xx 10^(-3)`

Answer

To find the initial concentration of acetic acid (CH₃COOH), we can use the ionization constant (Kₐ) and the concentration of hydrogen ions (H⁺) provided in the question. Here’s a step-by-step solution:

### Step 1: Write the ionization equation
The ionization of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Define initial concentrations
Let the initial concentration of acetic acid be \( C \). At the beginning (before ionization), we have:
- \([ \text{CH}_3\text{COOH} ] = C\)
- \([ \text{CH}_3\text{COO}^- ] = 0\)
- \([ \text{H}^+ ] = 0\)

### Step 3: Define changes at equilibrium
Let \( S \) be the amount that ionizes. At equilibrium, we have:
- \([ \text{CH}_3\text{COOH} ] = C - S\)
- \([ \text{CH}_3\text{COO}^- ] = S\)
- \([ \text{H}^+ ] = S\)

### Step 4: Use the given concentration of H⁺ ions
From the problem, we know:
\[ S = [\text{H}^+] = 3.4 \times 10^{-4} \, \text{M} \]

### Step 5: Write the expression for the ionization constant (Kₐ)
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{S \cdot S}{C - S} = \frac{S^2}{C - S} \]

### Step 6: Substitute known values into the Kₐ expression
Given \( K_a = 1.7 \times 10^{-5} \) and \( S = 3.4 \times 10^{-4} \):
\[ 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})^2}{C - 3.4 \times 10^{-4}} \]

### Step 7: Calculate \( S^2 \)
Calculate \( S^2 \):
\[ S^2 = (3.4 \times 10^{-4})^2 = 1.156 \times 10^{-7} \]

### Step 8: Rearranging the equation
Now we can rearrange the equation:
\[ C - 3.4 \times 10^{-4} = \frac{1.156 \times 10^{-7}}{1.7 \times 10^{-5}} \]

### Step 9: Calculate the right side
Calculating the right side:
\[ C - 3.4 \times 10^{-4} = 6.8 \times 10^{-3} \]

### Step 10: Solve for \( C \)
Now, solve for \( C \):
\[ C = 6.8 \times 10^{-3} + 3.4 \times 10^{-4} \]
\[ C = 6.8 \times 10^{-3} + 0.34 \times 10^{-3} \]
\[ C = 7.14 \times 10^{-3} \, \text{M} \]

### Final Answer
Thus, the initial concentration of CH₃COOH is approximately:
\[ C \approx 7.14 \times 10^{-3} \, \text{M} \]