Question #4d832

1 Answer
Mar 16, 2017

Answer:

#pOH=11.2#

Explanation:

We interrogate the equilibrium:

#2H_2O(l) rightleftharpoons H_3O^(+) + HO^(-)#

Now this equilibrium has been carefully measured under standard conditions, and at #298*K#:

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

This is a mathematical relationship, we can divide, multiply or otherwise manipulate. One thing we can do is to take #-log_10# OF BOTH SIDES,

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

But the RHS, #-log_(10)10^-14=-(-14)=14#, by the very definition of the #"logarithmic function"#. And thus,

#+14=-log_10[H_3O^+]-log_10[HO^-]#

And, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH#. This is simply by definition of the #p# function, and thus,

#+14=pH+pOH#

And thus we are given that #[H_3O^+]=1.55xx10^-3*mol*L^-1#, and so #pH=-(-2.81)=2.81.#

And so #pOH=11.19#

Capisce?