Calculate the pH at the equivalence point for the titration of 0.200 M methylamine (CH3NH2) with 0.200 M HCl. The Kb of methylamine is 5.0× 10–4.
I'm really confused. I've never done a problem like this before. Could you plug in the values for me? It just doesn't make any sense and my teacher won't explain it to me when I ask about it.
.........CH3NH3^+ + H2O ==> H3O^+ + NH3
I.........0.1M................0......0
C.........-x..................x......x
E.........0.1-x................x......x
Ka for CH3NH3^+ = (Kw/Kb for NH3) = (H3O^+)(NH3)/(O.1-x)
Solve for x = (H3O^+) and convert to pH.
But what's Kw? and the value for NH3? Is that just x? So should it be Kw/5.0x10-4 =(x)(x)/(0.1-x)
I typed too fast and made a typo. The equation should be
.......CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
Then Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.1-x)
Kw is the constant for H2O and at 25 C is 1E-14. Kb is given in the problem. Yes, Ka for CH3NH3^+ = 1E-14/5.0E-4 = ? There is no NH3--that's the typo. It should have been CH3NH2.
Ok, that's what I did before.
so 2E-11 = x^2/(0.1-x)
so x^2 = 2E-12 - 2E-11x
But what I don't understand then is how do we solve for x?
I used the quadratic equation but when I plugged in those values it gives me 0. So where am I going wrong?
I think you missed a sign.
2E-11 = x^2/(0.1-x)
x^2 + 2E-11x - 2E-12 = 0
I get 1.41E-6 and convert that to pH.