The degree of dissociation `(alpha)` of `CH_(3)COOH` solution having 0.1M concentration and pKa as 6, in presence of strong acid HCl having concentration 0.1M is `10^(-x)`. Then what is the value of x?

Explanation

To solve the problem, we need to find the degree of dissociation (α) of acetic acid (CH₃COOH) in the presence of a strong acid (HCl) and determine the value of x in the expression \(10^{-x}\).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Concentration of CH₃COOH = 0.1 M
- pKa of acetic acid = 6
- Concentration of HCl = 0.1 M

2. **Calculate Ka from pKa:**
\[ pKa = -\log(Ka) \implies Ka = 10^{-pKa} = 10^{-6} \]

3. **Set Up the Dissociation Reaction:**
The dissociation of acetic acid can be represented as:
\[ CH₃COOH \rightleftharpoons CH₃COO^- + H^+ \]
Let the degree of dissociation be α. Therefore, the concentration of dissociated acetic acid will be:
- Concentration of \(CH₃COO^-\) = \(0.1α\)
- Concentration of \(H^+\) from acetic acid = \(0.1α\)

4. **Consider the Contribution of HCl:**
Since HCl is a strong acid, it will completely dissociate:
\[ HCl \rightarrow H^+ + Cl^- \]
Thus, the concentration of \(H^+\) from HCl = 0.1 M.

5. **Total Concentration of H⁺ Ions:**
The total concentration of \(H^+\) ions in the solution will be:
\[ [H^+] = 0.1 + 0.1α \]

6. **Apply the Acid Dissociation Constant Expression:**
The expression for the acid dissociation constant \(Ka\) is given by:
\[ Ka = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]} \]
Substituting the known values:
\[ 10^{-6} = \frac{(0.1α)(0.1 + 0.1α)}{0.1(1 - α)} \]

7. **Simplify the Equation:**
Since α is very small (as indicated by the problem), we can approximate \(1 - α \approx 1\):
\[ 10^{-6} = \frac{0.1α(0.1 + 0.1α)}{0.1} \]
This simplifies to:
\[ 10^{-6} = α(0.1 + 0.1α) \]

8. **Neglect the α² Term:**
Since α is very small, we can neglect the \(0.1α^2\) term:
\[ 10^{-6} = 0.1α \]

9. **Solve for α:**
\[ α = \frac{10^{-6}}{0.1} = 10^{-5} \]

10. **Determine x:**
The problem states that \(α = 10^{-x}\). Therefore:
\[ 10^{-x} = 10^{-5} \implies x = 5 \]

### Final Answer:
The value of \(x\) is **5**.