In the laboratory, a general chemistry student measured the pH of a 0.322 M aqueous solution of methylamine, CH3NH2 to be 12.051. Use the information she obtained to determine the Kb for this base. Kb(experiment)

See explanation below

Explanation:

This is an equilibrium reaction of the methylamine when it's dissolved in water, therefore, we should write first this equilibrium:

CH₃NH₂ + H₂O <> CH₃NH₃⁺ + OH⁻

Now that we have the reaction and the innitial concentration of the base, we also have the pH, so we can calculate the final concentration of the methylamine and the oxydrilum ion.

As this is a base, we need to calculate the [OH⁻] concentration, and we need the pOH that is calculated with the expression:

14 = pH + pOH

From there, we solve for pOH:

pOH = 14 - 12.051 = 1.949

Now, we can calculate the concentration of OH:

[OH⁻] = antlog(-pOH)

[OH⁻] = antlog(-1.945) = 0.0112 M

Now that we have the concentration of OH, we can draw an ICE chart and from there, calculate the equilibrium constant Kb:

        CH₃NH₂ + H₂O <> CH₃NH₃⁺ + OH⁻     Kb = ?

i)         0.322                                  0             0

c)           -x                                     +x           +x

e)      0.322-x                                 x             x

Writting the Kb expression:

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]

and Replacing the values from the chart:

Kb = x²/0.322-x

But we already have the final concentration of OH, which will be the same of the CH3NH3+ and it's 0.0112 M, so replacing this value we finally will know the value of Kb:

Kb = (0.0112)² / (0.322 - 0.0112)

Kb = 4.036x10⁻⁴

And this is the equilibrium constant

2b2t

explanation:

2b2t

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