Problem: Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution:a solution that is 0.225 M in CH3NH2 and 0.120 M in CH3NH3Br

FREE Expert Solution

CH3NH2weak base → proton acceptor
H2O → will act as the weak acid → proton donor


Equilibrium reaction:        CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH-(aq)


Kb=productsreactantsKb=[CH3NH3H+][OH-][CH3NH2]

Solids and liquids are not included in the expression

4.4×10-4=[0.120+x][x][0.225-x]

Now, we need to determine if we can remove (–x) from the equation. To do so, we need to determine the ratio of the initial concentration and Kb

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4.4×10-4=[0.120+x][x][0.225-x]4.4×10-4(0.225-x)=0.120x+x20.225-x(0.225-x)9.9×10-5-4.4×10-4x=0.120x+x20=x2+0.12044x-9.9×10-5

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Problem Details

Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution:

a solution that is 0.225 M in CH3NH2 and 0.120 M in CH3NH3Br

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