Posted by Taylor on Tuesday, February 28, 2012 at 7:28pm.
Let's represent (C2H5O3N as B3N.
.............B3N + HOH ==> B3NH^+ + OH^-
initial......0.16...........0........0
change.......-x.............x........x
equil......0.16-x...........x........x
Kb = (B3BH^+)(OH^-)/(B3N)
Substitute Kb and from the ICE chart and solve for x = (OH^-). From that you can convert to pOH and to pH.
The second one is done the same way.
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