If #[H_3O^+]=4.5xx10^-3*mol*L^-1#, what are #pH#, and #pOH#, and #[HO^-]#?

1 Answer
May 19, 2017

Answer:

#pH=-log_(10)[H_3O^+]...............#

Explanation:

#(i)# #pH=-log_10(4.5xx10^-3)=-(-2.35)=2.35#.

For #pOH#, we recall that #14=pOH+pH# for protonolysis in water...

#(ii)# #pOH=-log_10(3.67xx10^-5)=-(-2.35)=4.44#; and thus, from the expression above, #pH=9.56#.

I will let you do the last problem.