`0.1M NaOH` is titrated with `0.1M, 20mL HA` till the point. `K_(a)(HA) = 6 xx10^(-6)` and degree of dissociation of `HA` is neglible (small) as compared to unity. Calculate the `pH` of the resulting solution at the end point [Use `log 6 ~~ 0.8]`

Explanation

To solve the problem of calculating the pH of the resulting solution at the endpoint of titrating 0.1M NaOH with 0.1M HA, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the number of moles of HA:**
\[ \text{Moles of HA} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 20 \, \text{mL} = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]
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