Question

What are [H+],[OH-],pH, and pOH of 0.120 M of CA(OH)2?

Answer 1

Well `pOH=-log_10[HO^-]`

Explanation:

And so here, `pOH=-log_10(0.240)=0.620`...why did I double the given concentration?

And if `pOH=0.620`, then `pH=14-0.620-=13.38`...a very basic solution....

Now clearly, `[Ca(OH)_2]=0.120*mol*L^-1` by specification...

`[HO^-]=0.240*mol*L^-1`...

Taking inverse logarithms...

`[H_3O^+]=10^(-13.38)*mol*L^-1=4.17xx10^-14*mol*L^-1`...