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

Explanation

To find the initial concentration of acetic acid (CH₃COOH) given the ionization constant (K) and the concentration of hydrogen ions (H⁺), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Ionization constant (K) of acetic acid (CH₃COOH) = \(1.7 \times 10^{-5}\)
- Concentration of hydrogen ions \([H^+]\) = \(3.4 \times 10^{-4}\)

2. **Set Up the Dissociation Equation:**
The dissociation of acetic acid can be represented as:
\[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]
Initially, let the concentration of acetic acid be \(C\) and the change in concentration due to dissociation be \(x\). At equilibrium, we have:
- \([H^+] = x\)
- \([CH_3COO^-] = x\)
- \([CH_3COOH] = C - x\)

3. **Substitute Known Values:**
From the problem, we know that:
\[ x = [H^+] = 3.4 \times 10^{-4} \]
Therefore, we can substitute \(x\) into the equilibrium expression.

4. **Write the Expression for K:**
The expression for the ionization constant \(K\) is given by:
\[ K = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]
Substituting the values we have:
\[ K = \frac{x \cdot x}{C - x} = \frac{x^2}{C - x} \]

5. **Plug in the Values:**
Substitute \(K\) and \(x\):
\[ 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})^2}{C - 3.4 \times 10^{-4}} \]

6. **Calculate \(x^2\):**
\[ (3.4 \times 10^{-4})^2 = 1.156 \times 10^{-7} \]
So the equation becomes:
\[ 1.7 \times 10^{-5} = \frac{1.156 \times 10^{-7}}{C - 3.4 \times 10^{-4}} \]

7. **Cross-Multiply:**
\[ 1.7 \times 10^{-5} (C - 3.4 \times 10^{-4}) = 1.156 \times 10^{-7} \]

8. **Distribute \(1.7 \times 10^{-5}\):**
\[ 1.7 \times 10^{-5}C - 5.78 \times 10^{-9} = 1.156 \times 10^{-7} \]

9. **Rearranging the Equation:**
\[ 1.7 \times 10^{-5}C = 1.156 \times 10^{-7} + 5.78 \times 10^{-9} \]
\[ 1.7 \times 10^{-5}C = 1.234 \times 10^{-7} \]

10. **Solve for \(C\):**
\[ C = \frac{1.234 \times 10^{-7}}{1.7 \times 10^{-5}} \approx 7.26 \times 10^{-3} \, \text{mol/L} \]

### Final Answer:
The initial concentration of CH₃COOH is approximately \(7.26 \times 10^{-3} \, \text{mol/L}\).