Ba(OH)2 ==> Ba^2+ + 2OH^-
Ksp = (Ba^2+)(OH^-)^2
(Ba^2+) = 6.6E-3 from Ba(OH)2 and 1E-2 from BaCl2.
Solve for (OH^-)
6.6×10−3 M Ba(OH)2 and 1.00×10−2 M BaCl2 what is [OH-] of solution?
2 answers
So do you add the 6.6E-3 and 1E-2 together as the value for Ba^2+?