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Solve an equilibrium problem (using an ICE table) to calculate the pH.

a solution that is 0.205 M in CH3NH2 and 0.110 M in CH3NH3Br.

i used th Ka of CH3NH2 and set up the problem as Ka=[H3O][CH3NH2]/[CH3NH3Br], but that was wrong. So i switched the equation to Ka=[H3O][CH3NH3Br]/[CH3NH2] but the pH i calculated from solving that equation was also wrong. Am I using a bad Ka or am I setting up the equation wrong?

The pH's i calculated were 3.63 and 3.09

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2 answers

  1. Frankly, I would use the Henderson-Hasselbalch equation. I assume your instructor has a reason for working the problem another way. Here is what I would do to follow the ICE method.

    CH3NH2 is a base. Think NH3. We write that equilibrium as
    NH3 + HOH ==> NH4^+ + OH^-
    CH3NH2 works the same way.
    CH3NH2 + HOH ==> CH3NH3^+ + OH^-

    These boards are difficult to make spaces so I must write the ICE chart as below; I suggest you redo it in the usual manner and write the I, C, and E amounts under the reactants and products. I think that will make it easier to see.
    Initial:
    CH3NH2 = 0.205 M
    CH3NH3^+ = 0.110 M
    OH^- = 0

    change:
    CH3NH2 = -x
    CH3NH2 = +x
    OH^- = +x

    equilibrium:
    CH3NH2 = 0.205-x
    CH3NH3^+ = 0.110+x
    OH^- = x

    Kb expression is
    Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
    Look up Kb and solve for x.
    When it comes to solving the equation, the x is small enough to ignore when combined with something else (0.205-x very nearly equals 0.205 and 0.110+x very nearly equals 0.110). I would then convert OH^- to pOH and subtract from 14 to obtain pH. I ran it quickly using the H-H equation and obtained 10.95 for the pH.

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  2. is these buffers solution NH3, NaC2H3O2
    NH3,NH4Cl
    HC2H3)2,HN)3
    KOH,KCl

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