Question

c=Calculate the OH- conc in a solution containing 6.2x10(-3)M Ba(OH)2 and 1.00x10(-2)M BaCl2?

I thought of using the BaCl2 conc to find the conc of Ba initially, then use that in the ice table for Ba(OH)2, and solve for OH. But i dnot think that's right.
Help? :/

Answer 1

I wonder if this problem expects us to use a Ksp for Ba(OH)2 or not. If not I see no reason to have included the concn of BaCl2. (And note that Ba(OH)2 is quite soluble--I've never even looked for a Ksp for it).(Another note: look in your text and see if Ksp is listed for Ba(OH)2 in Ksp tables. If so we should use it. If not, we don't use it. :-)]

Assume not a Ksp problem.
Then Ba(OH)2 --> Ba^2+ + 2OH^-
If [Ba(OH)2] = 0.0062M then OH^- is twice that.

If you think it is a Ksp problem then use the BaCl2 as a common ion and solve for OH^-. Repost if you get stuck and explain the problem.

Answer 2

Woooo! the easy way was correct c: Thanks man. But i still kinda don't know why you just doubled it. I thought it would havesomething to do witht he ice table and such since it was in the common ion+buffer chapter :s

Care explaining or?

Answer 3

You want an ICE chart?
.........Ba(OH)2 ==> Ba^2+ + 2OH^-
initial..0.0062M......0........0
change..-0.0062......0.0062...2*0.0062
equil....0..........0.0062M..0.0134M

But I didn't do all of that in my mind. I looked at Ba(OH)2, I see there are 2 OH^- per molecule of Ba(OH)2 and the problem tells me the [Ba(OH)2] = 0.0062M so I know OH^- is twice that.

Answer 4

Wow im an idiot XD thanks.
And i don't want to be a nuisance, but how would i do the next question?
0.320M (NH4)2SO4 and 0.492M NH3
(Still looking for OH- conc)

Would i find the H+ concentration from the double dissociation of the NH4compound? (i know how to find the oh from there)