When dealing with a buffer, we use the **Henderson-Hasselbalch equation**:

$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}\left(\frac{\mathbf{conjugate}\mathbf{}\mathbf{base}}{\mathbf{weak}\mathbf{}\mathbf{acid}}\right)}$

The reaction between HC_{2}H_{3}O_{2} and NaOH is shown below:

**HC _{2}H_{3}O_{2(aq)} + NaOH_{(aq)} **

(weak acid) (conjugate base)

***Na ^{+} is not included in the reaction because it is a neutral ion*

Calculate the initial amount of **HC _{2}H_{3}O_{2}** in moles:

$\overline{){\mathbf{molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{mol}}{\mathbf{L}}}\phantom{\rule{0ex}{0ex}}\mathbf{moles}\mathbf{}{\mathbf{HC}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\frac{\mathbf{mol}\mathbf{}{\mathbf{HC}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}}{\overline{)\mathbf{L}}}\mathbf{)}\mathbf{\times}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{L}}$

**moles HC _{2}H_{3}O_{2} = 2.0 mol HC_{2}H_{3}O_{2}**

We need to construct an ICF chart to determine the amounts of the species after the reaction:

Substitute values in the Henderson-Hasselbalch equation:

$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}\left(\frac{\mathbf{conjugate}\mathbf{}\mathbf{base}}{\mathbf{weak}\mathbf{}\mathbf{acid}}\right)}\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\left(\frac{\mathbf{x}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{x}}\right)\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{-}{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{log}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{x}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{0}}{\mathbf{log}}\mathbf{=}\frac{\overline{)\mathbf{log}}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{x}}\mathbf{\right)}}{\overline{)\mathbf{log}}}$

***log is a base of 10, when you divide a number by log, you will take 10 raised to the power of the number*

What quantity (moles) of NaOH must be added to 1.0 L of 2.0 M HC_{2}H_{3}O_{2} to produce a solution buffered at each pH?

a. pH = pK_{a}

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