What is #pH# of an aqueous solution that is #0.016*mol*L^-1# with respect to #NaOH#?

1 Answer
Apr 5, 2017

Answer:

#pH=12.2#

Explanation:

In aqueous solution, #pH=-log_(10)[H_3O^+]#, and likewise, #pH=-log_(10)[HO^-]#.

Now since we interrogate the equilibrium:

#2H_2Orightleftharpoons H_3O^+ + ""^(-)OH# where #K_"water"=10^(-14)#, then..........

#[H_3O^+][HO^-]=10^(-14)#, and we take #log_10# of both sides......

#log_10[H_3O^+]+ log_10[HO^-]=log_(10)10^(-14)=-14#

On rearrangement,

#14=-log_10[H_3O^+]-log_10[HO^-]=pH+pOH#.

And this is our defining relationship, #pH+pOH=14# (and this has to be known and used!)

So, we find the #pOH# of a solution that is #0.016*mol*L^-1# with respect to #NaOH#:

#pOH-=-log_(10)(0.016)=1.80#

And finally,

#pH=14-pOH=14-1.80=12.2#

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