How do I find #pH# and #pOH# for solutions of aqueous hydrogen fluoride or sodium fluoride?

1 Answer
Nov 21, 2016

Answer:

You have the measurements, but you must use the definitions of #pH# and #pOH#. On the basis of the data given, you cannot measure #pH# for the basic sodium fluoride solution.

Explanation:

Water dissociates according to the following equilibrium:

#2H_2O rightleftharpoonsH_3O^+ +HO^-#

Now it is well known that for a given temperature, the extent of this autoprotolysis reaction is a constant:

#K_w=([H_3O^+][HO^-])/([H_2O]^2)#

And we can simplify this to
#K_w=[H_3O^+][HO^-]# because #[H_2O]# is effectively constant.

By careful measurement we know that #K_w=10^(-14)# at #298*K#

Now this is a mathematical expression, which we are free to manipulate, provided that we do it to BOTH sides of the equation. Now we can take #log_10# of both sides of the equation:

#log_10K_w# #=# #log_10[H_3O^+] +log_10[""^(-)OH]#

But by definition #-log_10[H_3O^+]=pH# and #-log_10[HO^-]=pOH#.

And also by definition #pK_w=-log_10K_w# #=-log_(10)10^(-14)=14.#

So we get our defining relationship:

#pH+pOH=14#, under the given conditions of your question. You should commit this relationship to memory. You won't be asked to derive it at A level.

So get an electronic calculator, and have a go at your table. You will need #K_a# for hydrofluoric acid, #=10^(-3.17)#.