How do you calculate the pH of a .40 solution of Ba(OH)_2Ba(OH)2 when 25.0 mL is added to 250 mL of water?

1 Answer
anor277
Jul 20, 2017

Well, pH=12-13.pH=12−13.

Explanation:

We first calculate [Ba(OH)_2]="Moles of stuff"/"Volume of solution"[Ba(OH)2]=Moles of stuffVolume of solution

=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)=0.40⋅mol⋅L−1×25.0×10−3⋅mol⋅L−1275.0×10−3⋅L

=0.0364*mol=0.0364⋅mol with respect to [Ba(OH)_2][Ba(OH)2], i.e. 0.07272*mol*L^-10.07272⋅mol⋅L−1 with respect to [HO^-][HO−].

pOH=-log_10[HO^-]=-log_10(0.07272)=1.14pOH=−log10[HO−]=−log10(0.07272)=1.14

pH=14-pOH=14-1.14=12-13pH=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*mL250⋅mL rather than when 25*mL25⋅mL is added to 250*mL250⋅mL water......

Related questions
See all questions in pH calculations
Impact of this question
1763 views around the world
You can reuse this answer
Creative Commons License