Question

What am I doing wrong in this titration problem?

Calculate the ph at the equivalence point for the following 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 x 10^-11) to get 2.3x10^-12, which is wrong.

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

Answer 1

Equivalence point:

Mol acid = Mol Base

M1*V1 = M2*V2

Assume V1 = V2 since there is no extra data, they should be in the same solution

M1 = M2 which makes sens since 0.2 = 0.2

In equilibrium

CH3NH2 + HCl ---> CH3NH3+ and Cl-

This will form an equilibrium

CH3NH3+ and H2O <---> H3O+ and CH3NH2

Recall the equilibrium

Ka = [H3O+][CH3NH2]/[CH3NH3+]

Since you have H3O+ ions, expec acidic solution

Ka = 2.3*10^-11

[H3O+] = [CH3NH2 ] = x

[CH3NH3] = 0.2 - x

Substitute in Ka expression

2.3*10^-11 = x*x / (0.2-x)

As you said, assume "-x" is so samll compared to 0.2

2.3*10^-11 = x^2 / 0.2

(2.3*10^-11)*0.2 ) = X^2

4.6*10^-12 = x^2

x = sqrt(4.6*10^-12) = 2.14*10^-6

NOTE: I don't know why you divided 0.2 by two... may be because the volume increased by 2 (i.e. 25 ml of acid and 25 ml of base, when adding will be 50 ml, recalculating concentrations you will have a diluted concnetration by a factor of 2x)

NOTE2: you are forgetting to take the square root of 2.3x10^-12

I mean: 2.3x10^-12 = x^2 as you said... you are forgetting to solve for x...

For your value... I got = 1.51*10^-6 which is exactly the value you got

NOTE: since this is exactly the value, I'm pretty sure you have same volumes of acid and base, so you have a double volume... thats why you are dividing by 2... its the only thing that makes sense to me

good luck, please let me know if oyu need more help