Problem: A 120.0 -mL buffer solution is 0.110 M in NH3 and 0.135 M in NH4Br. What mass of HCl can this buffer neutralize before the pH falls below 9.00?        

Determine the moles of the weak acid and conjugate base:

$\mathbf{moles}\mathbf{ }{\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{120}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\left(\frac{{10}^{-3} \cancel{L}}{1 \cancel{\mathrm{mL}}}\right)\left(\frac{0.110 \mathrm{mol}}{1 \cancel{L}}\right)\mathbf{=}$0.0132 mol

$\mathbf{moles}\mathbf{ }{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{=}\mathbf{120}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\mathbf{\left(}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{135}\mathbf{ }\mathbf{mol}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{L}}}\mathbf{\right)}\mathbf{=}$0.0162 mol

Construct an ICF Chart:

$\boxed{\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)}}$

${\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{14}\mathbf{-}{\mathbf{pK}}_{\mathbf{b}}$

${\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{14}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{75}\mathbf{=}$9.25

$$\begin{gathered} \mathbf{9}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{25}\mathbf{+}\mathbf{log}\mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)} \\ \mathbf{log}\mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)}\mathbf{=}\mathbf{9}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{25} \\ \mathbf{log}\mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{25} \end{gathered}$$

$$\begin{gathered} \mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{25}} \\ \mathbf{\left(}\frac{{\mathbf{NH}}_{\mathbf{3}}}{{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5623} \\ \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0132}\mathbf{-}\mathbf{x}}{\mathbf{0}\mathbf{.}\mathbf{0162}\mathbf{+}\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5623} \end{gathered}$$

$$\begin{gathered} \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0132}\mathbf{-}\mathbf{x}}{\mathbf{0}\mathbf{.}\mathbf{0162}\mathbf{+}\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5623} \\ \mathbf{0}\mathbf{.}\mathbf{5623}(0.0162+x)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0132}\mathbf{-}\mathbf{x} \\ \mathbf{9}\mathbf{.}\mathbf{10926}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{5623}\left(x\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0132}\mathbf{-}\mathbf{x} \\ \mathbf{0}\mathbf{.}\mathbf{5623}\left(x\right)+x=\mathbf{0}\mathbf{.}\mathbf{0132}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{10926}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}} \\ \frac{\cancel{\mathbf{1}\mathbf{.}\mathbf{5623}}\mathbf{x}}{\cancel{\mathbf{1}\mathbf{.}\mathbf{5623}}}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{09074}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{1}\mathbf{.}\mathbf{5623}} \end{gathered}$$