# 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 of the solution to be 11.77.
(a) Write the expression for the equilibrium constant, Kb, for methylamine.
(b) Calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution.
(c) Calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established.
The 50.0 mL sample of the methylamine solution is titrated with HCl solution of unknown concentration. The equivalence point of the titration is reached after a volume of 36.0 mL of the HCl solution is added. The pH of the solution at the equivalence point is 5.98.
(d) Write the net-ionic equation that represents the reaction that takes place during the titration.
(e) Calculate the concentration of the HCl solution used to titrate the methylamine.
(f) using the axes provided, sketch the titration curve that results from the titration described above. On the graph, clearly label the equivalence point of the titration.

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1. a) Kb=[CH3NH3][OH-]/[CH3NH2][H20]
b) -log[H+]=pH=11.77
pH+pOH=14.00
14.00-11.77=2.23=pOH
[OH-]=10^-pOH=10^(-2.23)=.0058M

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2. Everything Christina did is correct.
Starting with C
4.4E-9=[OH-]^2/x x=.078

d. Ch3Nh2 + H+ -> CH3NH3+

e.(30)(x)=(50)(.078) since moles are equal at equivalence.
x=.13 Molar

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3. Why is the [] of the CH3NH3 the same as the [OH-]?

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4. in your equilibrium constant expression do not include water because it is a liquid

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5. 142

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