# 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) ${\mathbf{K}}_{{\mathbf{b}}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[{\mathrm{CH}}_{3}{\mathrm{NH}}_{3}{H}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{NH}}_{2}\right]}$

Solids and liquids are not included in the expression

$\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{\left[0.120+x\right]\left[x\right]}{\left[0.225-x\right]}$

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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$\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{\mathbf{\left[}\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{+}\mathbf{x}\mathbf{\right]}\mathbf{\left[}\mathbf{x}\mathbf{\right]}}{\left[0.225-x\right]}\phantom{\rule{0ex}{0ex}}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\left(0.225-x\right)\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}{\overline{)0.225-x}}\overline{)\left(0.225-x\right)}\phantom{\rule{0ex}{0ex}}\mathbf{9}\mathbf{.}\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{0}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{12044}\mathbf{x}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}$

81% (132 ratings) ###### 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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