How do you calculate #[OH^-]# in #[H_3O^+] = 2.1 times 10^-8# #M# at 25°C?

1 Answer
Mar 29, 2017

Answer:

How do we calculate #[HO^-]# IF #[H_3O^+]# #=2.1xx10^-8*mol*L^-1#?

Well, #[""^(-)OH]=4.78xx10^-7*mol*L^-1.#

Explanation:

We know that #K_w=[H_3O^+][""^(-)OH]=10^-14#, in water at #298*K#.

We take #log_10# of both SIDES............

#log_10K_w=log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14#

But #log_(10)10^-14=-14# by definition............

And if we DEFINE,
#-log_10K_w=pK_w#, and #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH#

Then................

#pH+pOH=pK_w=14#.

You should commit this last result to memory, as it makes the treatment of acidity/basicity straightforward.

So if #[H_3O^+]=2.1xx10^-8*mol*L^-1#, #pH=-(-log_(10)2.1xx10^-8)=-(-7.68)=+7.68#

And thus #pOH=14-7.68=6.32#, and #[HO^-]=10^(-6.32)=4.78xx10^-7*mol*L^-1.#

The solution is slightly BASIC....................