Question

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

Answer 1

`[H_3O^+]=10^(-1.90)*mol*L^-1=1.26xx10^-2*mol*L^-1`

Explanation:

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

`2H_2O(l)rightleftharpoonsH_3O^+ + HO^-`

WE know by precise measurement at `298*K` under standard conditions, the ion product......

`K_w=[H_3O^+][HO^-]=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{K_w}=log_10{[H_3O^+][HO^-]}=log_10(10^-14)`

And thus `log_10[H_3O^+]+log_10[HO-]=-14`, and on rearrangement.....

`14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)`

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

`pH+pOH=14`

We have `pH=1.90`. And thus we take antilogs.......

`[H_3O^+]=10^(-1.90)*mol*L^-1=1.26xx10^-2*mol*L^-1`

Back in the day, instead of using calculators to find logs etc, students would have to use logarithmic tables to find `pH` 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.