6.6×10−3 M Ba(OH)2 and 1.00×10−2 M BaCl2 what is [OH-] of solution?

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  1. 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^-)

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  2. So do you add the 6.6E-3 and 1E-2 together as the value for Ba^2+?

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