`20mL` of `0.2M NaOH` are added to `50mL` of `0.2M` acetic acid `(K_(a)=1.85xx10^(-5))`
Take `log2=0.3,log3=0.48`
(1) What is `pH` of solution?
(2) Calculate volume of `0.2M NaOH` required to make the `pH` of origin acetic acid solution.`4.74`.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part 1: Calculate the pH of the solution after mixing NaOH and acetic acid.

1. **Calculate the number of moles of NaOH and acetic acid:**
- Moles of NaOH = Molarity × Volume (in L)
\[
\text{Moles of NaOH} = 0.2 \, \text{M} \times 0.020 \, \text{L} = 0.004 \, \text{mol} \, (4 \, \text{mmol})
...