# Why do strong acids have a low pH?

Mar 18, 2018

Because of the way we express the $p$ function....

#### Explanation:

By definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$. And the use of the logarithmic function dates back to pre-electronic calculator days, when students, and engineers, and scientists, used logarithmic tables for more complex calculations, the which a modern calculator, available for a dollar or so, would EAT today....

For a strong acid, say $H C l$ at MAXIMUM concentration, approx. $10.6 \cdot m o l \cdot {L}^{-} 1$, which is conceived to ionize completely in aqueous solution, we gots...

$H C l \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + C {l}^{-}$

Now here, $\left[{H}_{3} {O}^{+}\right] = 10.6 \cdot m o l \cdot {L}^{-} 1$....

And so $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left\{10.6\right\} = - \left(+ 1.03\right) = - 1.03$..

And thus for stronger acid $\left[{H}_{3} {O}^{+}\right]$ GIVE A MORE NEGATIVE $p H$....

For background...

Just to note that in aqueous solution under standard conditions, the ion product...

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$...

And we can take ${\log}_{10}$ of both sides to give....

${\log}_{10} {K}_{w} = {\log}_{10} {10}^{- 14} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$.

And thus.... $- 14 = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$

Or.....

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

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

$14 = p H + p O H$

By definition, $- {\log}_{10} \left[{H}^{+}\right] = p H$, $- {\log}_{10} \left[H {O}^{-}\right] = p O H$.