Calculate the `[H^(o+)]` and `[overset(Theta)OH]` of `0.0315g` of `HNO_(3)` in `500 mL` of water. Calculate `pH` and `pOH` also.

Answer

A, C

Explanation

`(Mw_(2)` of `HNO_(3) = 53 g mol^(-1))`
`M_(HNO_(3)) = (W_(2) xx 1000)/(Mw_(2) xx "vol of sol. In mL")`
`= (0.0315 xx 1000)/(63xx500) = 0.001= 10^(-3)M`.
`[H^(o+)] = 10^(-3)M,pH = 3`
`[overset(Theta)OH] = (K_(w))/([H^(o+)]) = (10^(-14))/(10^(-3)) = 10^(-11)M`
`pOH = 11`.