How do you calculate the pH of a .40 solution of #Ba(OH)_2# when 25.0 mL is added to 250 mL of water?

1 Answer
Jul 20, 2017

Answer:

Well, #pH=12-13.#

Explanation:

We first calculate #[Ba(OH)_2]="Moles of stuff"/"Volume of solution"#

#=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)#

#=0.0364*mol# with respect to #[Ba(OH)_2]#, i.e. #0.07272*mol*L^-1# with respect to #[HO^-]#.

#pOH=-log_10[HO^-]=-log_10(0.07272)=1.14#

#pH=14-pOH=14-1.14=12-13#.

Note that you might have been a bit careless with the wording of your question. You get a different answer when the solution is diluted to #250*mL# rather than when #25*mL# is added to #250*mL# water......