The dissociation of the weak acid, nitrous acid, $HN{O}_2,$ takes place according to the reaction:

$HN{O}_2(aq)H^{+}(aq)+N{O}_{2-(aq)}^{-}K=7\mathrm{.}2{10}^{-4}$

When $1\mathrm{.}00$ mole of $HN{O}_2$ is added to $1\mathrm{.}00L$ of water, the $H+$ concentration at equilibrium is $0\mathrm{.}0265M$

A$)$ Calculate the value of $Q$ if $1\mathrm{.}00L$ of water is added?

B$)$ How will reaction shift if $1\mathrm{.}00L$ of water is added?