# The pH of a lime is 1.90. What is the [H_3O^+]?

Jul 30, 2017

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 1.90} \cdot m o l \cdot {L}^{-} 1 = 1.26 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We know that water undergoes autoprotolysis according to the following equation......

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

WE know by precise measurement at $298 \cdot K$ under standard conditions, the ion product......

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

Now this is an equation, that I may divide, subtract, multiply, PROVIDED that I does it to both sides. One thing I can do is to take ${\log}_{10}$ of BOTH SIDES........

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

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

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

And so (FINALLY) we gets our working relationship, which you should commit to memory.....

$p H + p O H = 14$

We have $p H = 1.90$. And thus we take antilogs.......

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 1.90} \cdot m o l \cdot {L}^{-} 1 = 1.26 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

Back in the day, instead of using calculators to find logs etc, students would have to use logarithmic tables to find $p H$ etc. A modern electronic calculator, and I picked up mine for £1-00 in a remainders shop, is so much easier.

Note that most things that taste good and piquant, limes, lemon, wine, cheese, butter, are acidic.