Gen Chem help! Destroyer problem

[calicarbocation]

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This is number 89 out of the gen chem section of the destroyer:

The Kb for C2H5NH2 is 6 x 10^-4. Calculate the pH of a 2.0 M solution at 25 degrees C.

The answer came out to be 12.5....So we need to be able to do this kind of math in our heads without a calculator right? I just need some advice of how to walk through the math in my head with all the divisions and logarithms, etc with it comes to Ka/Kb/pH/pKa problems. Anyone else have trouble with this?

 

[Skateguitar]

easy way to do this:
[OH] = sqrt[(Molarity)(Kb)]
or for an acid [H] = sqrt[(Molarity)(Ka)]

The question is asking for the Ph, However we are given the K of a base (Kb) so right away you know you must get the POH and then convert it to Ph.

Lets plug in: [OH] = sqrt[(2M)(6x10^-4)]
[OH] = sqrt[12x10^-4]

(the squareroot of 12 is around 3.5, an estimation. The square root of x10^-4 is just half that x10^-2)

[OH] = 3.5x10^-2
Notice this is the OH concentration. not the POH or the pH, we must convert to those. Also, if we had gotten 1x10^-2, then the POH would have been 2. However, since 3.5 is bigger than 1, 3.5x10^-2 is bigger which should lower the POH because as the OH goes up, the POH goes down (i am not good at explaining that, sorry, maybe someone else can help) so we estimate the POH to 1.5.

POH + Ph = 14
1.5 + Ph = 14
Ph = 14 - 1.5
Ph = 12.5