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

1 Answer
Jul 30, 2017

Answer:

#[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.