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

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

Kb=productsreactantsKb=

Solids and liquids are not included in the expression

4.4×10-4=<0.120+x><0.225-x>

Now, we must identify if we have the right to remove (–x) from the equation. To perform so, we need to identify the proportion of the initial concentration and Kb:

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4.4×10-4=<0.120+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 trouble (using an ICE table) to calculate the pH of each solution:

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

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