Find the $pH$ of a $7.6\cdot 10^{-4}\;M$ solution of $HNO_3$ ! First we write down the equation of it's dissociation. Is it..
a) $HNO_3 (aq)$ 
$NO_2^+ (aq)$ $+$ $OH^- (aq)$
b) $HNO_3 (aq)$ $+$ $OH^- (aq)$ $NO_3^- (aq)$ $+$ $H_2O(l)$
c) $HNO_3 (aq)$ $H^+ (aq)$ $+$ $NO_3^- (aq)$
d) $HNO_3 (aq)$ $+$ $H^+$ $NO_2^+ (aq)$ $+$ $H_2O(l)
?
c)$ HNO_3 (aq)$ $H^+ (aq)$ $+$ $NO_3^- (aq)$
This reaction is complète. True or false ?
True, more exactly: $HNO_3 (aq)$ $\longrightarrow$ $H^+ (aq)$ $+$ $ NO_3^- (aq)$
What would be $[H^+(aq)]$?
$[H^+(aq)]$ will have the same molar concentration than $HNO_3$, so: $[H^+(aq)]$ $=$ $7.6 \cdot 10^{-4}\;M$
Calculate the $pH$!
$pH$ $=$ $-log(7.6 \cdot 10^{-4})$ $=$ $3.12$