Ba(OH)2 ionizes 100%; therefore,
..........Ba(OH)2 ==> Ba^2+ + 2OH^-
I.........0.00988.....0.........0
C.......-0.00988....0.00988..2*0.00988
E.........0.........0.00988..0.0198
.........H2O --> H^+ + OH^-
I........liquid...0....0.0198
C........liquid...x.....x
E........liquid...x...0.0198+x
I assume this is standard Kw.
Kw = (H^+)(OH^-)
Substitute the E line into Kw and solve for x = (H^+)
You can make the assumption that 0.0198+x = 0.0198
C.......
Calculate the hydrogen ion concentration, [H+], in 0.00988 M Ba(OH)2. Careful!
1 answer
1 answer