What is #[HO^-]# in an aqueous solution for which #[H_3O^+]=3.2xx10^-3*mol*L^-1#?

1 Answer
anor277
Jul 30, 2017

#[HO^-]=3.16xx10^-12*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 #[H_3O^+]=3.2xx10^-3*mol*L^-1#. #pH=-log_10(3.2xx10^-3)=2.50#. And thus #pOH=11.50#.

And now we take antilogs.....#[HO^-]=10^(-11.50)*mol*L^-1#

#=3.16xx10^-12*mol*L^-1# as required............

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