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What am I doing wrong in this titration problem? Calculate the ph at the equivalence point...

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Chemistry

What am I doing wrong in this titration problem?

Calculate the ph at the equivalence point for thefollowing titration 0.20M HCl versus 0.20M methylamine (CH3NH2).The Ka of methylammonium is 2.3x10^-11.

First I have to divide .20M methylamine by 2(Why?) to get .10M

Then, I set up the equilibrium:

(2.3 x 10^-11) = x^2 / .10M

Since the ka is SO small, I just multiplied .10 with (2.3 x10^-11) to get 2.3x10^-12, which is wrong.

Why is this wrong? Since the Ka is small, the approxiamationmethod should work and I won't need to do the quadratic. Instead, Iam told that the x value is 1.5x10^-6 from the quadratic.

Answer & Explanation Solved by verified expert
4.2 Ratings (828 Votes)
Equivalence point Mol acid Mol Base M1V1 M2V2 Assume V1 V2 since there is no extra data they should be in the same solution M1 M2 which makes sens since 02 02 In equilibrium CH3NH2 HCl CH3NH3 and Cl This will form an equilibrium CH3NH3    See Answer
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