Asked by candace
I have this on my homework but cannot fine any examples in the book or my notes that relate to a problem set up this way, can you help?
What are the concentrations of OH– and H in a 0.00078 M solution of Ba(OH)2 at 25 °C? Assume complete dissociation.
What are the concentrations of OH– and H in a 0.00078 M solution of Ba(OH)2 at 25 °C? Assume complete dissociation.
Answers
Answered by
DrBob222
[Ba(OH)2] = 7.8E-4M
......Ba(OH)2 ==> Ba^2+ + 2OH^-
I....7.8E-4.......0........0
C....-7.8E-4...7.8E-4...2*7.8E-4
E.....0........7.8E-4..1.56E-3
So (OH^-) = 1.56E-3M and use
(H^+)(OH^-) = Kw = 1E-14 to solve for (H^+).
......Ba(OH)2 ==> Ba^2+ + 2OH^-
I....7.8E-4.......0........0
C....-7.8E-4...7.8E-4...2*7.8E-4
E.....0........7.8E-4..1.56E-3
So (OH^-) = 1.56E-3M and use
(H^+)(OH^-) = Kw = 1E-14 to solve for (H^+).
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