If #[HO^-]=6.0xx10^-10*mol*L^-1#, what is #pH# for this solution?

1 Answer
Oct 31, 2016

Answer:

#[H^+] ("or "[H_3O^+])=10^(-4.78)*mol*L^-1#, #pH=4.78#, #pOH=9.22#

Explanation:

In aqueous solution at #298*K#, water undergoes the following equilibrium reaction:

#H_2O(l) rightleftharpoons H^(+) + HO^(-)#

Alternatively, and perhaps now more commonly we would write:

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

This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:

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

If we take negative logarithms to the base 10 of both sides we get:

#pK_w=14=pOH+pH#, where the #pH# functions means #-log_10[H_3O^+]#, etc.

We are given #[HO^-]=6.0xx10^-10#, so #pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22#, and thus #pH=14-9.22=4.78#.

Is this solution acidic or basic?

The given reaction occurs at #298*K#. Given that this is a bond-breaking reaction, how do you think #pK_w# would evolve at #323*K# or #373*K#?