Step 1

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{L}}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$**= 0.50 mol NH**_{3}

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{L}}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$**= 0.20 mol NH _{4}Cl**

Step 2

**mol NH _{4}^{+} = 0.20 mol - 0.110 mol = 0.09 mol**

**mol NH _{3} = 0.50 mol + 0.110 mol = 0.61 mol**

Consider a buffer solution that is 0.50 M in NH_{3} and 0.20 M in NH_{4}Cl. For ammonia, pK_{b}= 4.75. Calculate the pH of 1.0 L upon addition of 0.110 mol of solid NaOH to the original buffer solution.

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