Solution of `CH_3 COOH [Ka = 10^(-5) " at " 25^@ ] ` is been neutralized by using NaOH solution. pH of the solution is x at 20% neutralization and it is y at 80% neutralization then value of y-x will be `[ log_(10)^(2) = 0.30 ]`

Explanation

To solve the problem, we need to determine the pH changes during the neutralization of acetic acid (CH₃COOH) with sodium hydroxide (NaOH) at two different levels of neutralization: 20% and 80%.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The dissociation constant (Ka) for acetic acid (CH₃COOH) is given as \(10^{-5}\).
- The neutralization occurs at 20% and 80% completion.
- We need to find the difference in pH (y - x) at these two levels of neutralization.
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