- #1
mesa
Gold Member
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Homework Statement
Calculate the pH of a solution that is .20M NH3 and .35M NH4Cl with a Kb(NH3)=1.8x10^-5
The Attempt at a Solution
So first calculated the M of OH- produced by the NH3,
[OH-]=√((1.8x10^(-5))x.20) = 1.9x10^-3
Since NH4Cl will dissociate in H2O it will leave NH4+, NH4+ + H2O ---> NH3 + H30+
Ka would be [1x10^(-14)]/[1.8x10^(-5)] = 5.6x10^(-10) giving us a rather insignificant amount of hydronium produced compared to the amount of OH- from the NH3 but let's figure it none the less,
[H3O] = √(5.6x10^(-10))x.35 = 1.4x10^(-5)
Since the hydronium will react with the OH- subtracting the two gives us 1.88x10^-3M OH-
giving us a pOH of 2.73, or pH of 11.27
The answer is pH 9.01 so I am waaaaay off.
This is new stuff for my brain and that's the excuse I'm sticking with :)