12.25

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Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

12.25

Postby Rakhi Ratanjee 1D » Thu Dec 07, 2017 4:30 pm

Q: Calculate the initial molarity of Ba(OH)2 and the molarities of Ba+2, OH-, and H30+ in an aqueous solution that contains 0.43 grams of Ba(OH)2 in 0.100L of solution.

For this question, I understand how to find the molarity, but I'm confused on how to set up the dissociation equation. How do you know when to react the acid or base with water to form a conjugate acid/conjugate base pair and when you just have to make it the dissociation of the acid/base into the ions?
The way I approached the problem was to set up the Kb as x^3/(concentration of Ba(OH)2). Why won't this method work?

Cooper1C
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 12.25

Postby Cooper1C » Thu Dec 07, 2017 6:43 pm

I believe the acid/base will not dissociate unless it is in some sort of solvent. In this case, the problem states Ba(OH)2 is in an aqueous solution.

Ba(OH)2 is a strong acid, so you do not need to set up an ice box, and you are not given Kb. Because Ba(OH)2 is a strong base, we assume it is 100% ionized at equilibrium, and [Ba2+] = 2*[OH-] = the initial concentration of Ba(OH)2, and you calculate [H3O+] from [OH-].


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