A $100 \; \text{mg}$ of $\text{HNO}_{3}$ (strong acid) is added to water, bringing the final volume to $1.0 \; \text{liter}.$ Consider the atomic weights of $\text{H, N,}$ and $\text{O,}$ as $1 \; \text{g/mol,} \; 14 \; \text{g/mol,}$ and $16 \; \text{g/mol,}$ respectively. The final $\text{pH}$ of this water is
(Ignore the dissociation of water.)
- $2.8$
- $6.5$
- $3.8$
- $8.5$