Respuesta :

loneguy

[tex]\text{HNO}_3 (aq) \longrightarrow \text{H}^{+} + \text{NO}_3^{-} (aq)\\\\\text{Here ,} ~~ [ \text{H}^{+} ] = 0.056~ M\\\\\\\text{pH} = -\log[\text{H}^+}] = -\log(0.056) = 1.252[/tex]