pH = 11.48
pOH = 2.52
pOH = -log(OH^-); therefore, OH^- = 0.003 but that's an estimate. You should do it more accurately.
........NH3 + H2O ==> NH4^+ + OH^-
I......0.5.............0.......0
C.......-x.............x.......x
E......0.5-x...........x.......x
x = 0.003 from above.
Substitute into the Kb expression for NH3 and solve for Kb.
calculate the kb of ammonia if the ph of .5M solution is 11.48
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