If pH=1.65, what is [H_3O^+]?

Sep 13, 2017

$\left[{H}^{+}\right]$ or $\left[{H}_{3} {O}^{+}\right] = 0.0224 \cdot m o l \cdot {L}^{-} 1$

Explanation:

For the equilibrium,

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$...

alternatively, ${H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + H {O}^{-}$

${K}_{\text{eq}} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions.....

And we take ${\log}_{10}$ of BOTH SIDES, and gets....

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = {\log}_{10} \left({10}^{-} 14\right) = - 14$

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{p H +} {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{p O H} = + 14$

And so $p H + p O H = 14$, under standard conditions, the which we assume.

And thus for ${\left[H\right]}^{+}$/$\left[{H}_{3} {O}^{+}\right]$, we take ${10}^{- 1.65} \cdot m o l \cdot {L}^{-} 1 = 0.0224 \cdot m o l \cdot {L}^{-} 1$.

What is $\left[H {O}^{-}\right]$ for this solution?