Analytical Chemistry

posted by .

Calculate the pH at the equivalence point for the titration of 0.160 M methylamine (CH3NH2) with 0.160 M HCl. The Kb of methylamine is 5.0× 10–4.

Methylamine is a weak base and reacts with HCl to give the methylammonium ion.

HCl + CH3NH2 <---> CH3NH3^+ + Cl^-
0.160M 0.160M 0 0
-x -x +x +x
0.160-x 0.160-x x x

kb= x^2/(0.160-x)(0.160-x)

Have I set up the ice chart correctly? I'm not sure.

  • Analytical Chemistry -

    I apologize that the ice chart got smushed like this when i posted the problem.

  • Analytical Chemistry -

    I think you should have

    CH3NH3 + H2O----> H3O+ + CH3NH2


    Ka=Kw/kb= x^2/(0.160-x)

    Sqrt*(Ka*(0.160))=H+

    ph=-log[H+]

  • Analytical Chemistry -

    I did not get the right answer by this method.

  • Analytical Chemistry -

    Yes, because the volume doubles at the eq. pt, and the the concentration would change (i.e., the M would be cut in half). I didn't put in the right concentration when I gave you the set up.


    Notice what I changed.

    CH3NH3 + H2O----> H3O+ + CH3NH2


    Ka=Kw/kb= x^2/(0.08-x)

    Sqrt*(Ka*(0.08))=H+

    ph=-log[H+]

    I apologize about that!!!!

  • Analytical Chemistry -

    its alright. I figured out the answer. thanks for your help! :)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    A few questions I don't really get and need to see the work for A 50.0 mL sample of 0.55 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.51 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 …
  2. AP Chemistry

    In the titration of 77.5 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 10-4), with 0.38 M HCl, calculate the pH under the following conditions. (a) after 50.0 mL of 0.38 M HCl has been added (b) at the stoichiometric point
  3. Chemistry!!

    Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0× 10^–4.
  4. Chemistry

    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.
  5. chemistry

    CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH …
  6. AP Chemistry

    CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH …
  7. Chemistry II

    Calculate the pH at the equivalence point in the titration of 50ml of 0.20 M methylamine (Kb=4.3*10^-4) with a 0.40 M HCL solution.
  8. Chemistry

    Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0× 10–4.
  9. General Chemistry

    1,When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a soHowever, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble [Pb(OH)4]2–(aq) complex ion.lid precipitate …
  10. CHemistry

    Calculate the pH at the equivalence point for the titration of 0.140 M methylamine (CH3NH2) with 0.140 M HCl. The Kb of methylamine is 5.0× 10–4.

More Similar Questions