# Buffers

Buffers are solutions that contain appreciable amounts of acid and conjugate base (or base and conjugate acid). If the ratio of acid to conjugate base concentration (or base to conjugate acid concentration) lies between 0.1 and 10, you have a buffer!

Buffers have the ability to resist a change in pH.

## Henderson-Hasselbalch Equation

The pH of a buffer solution can be determined without resorting to ICE tables and solving an equilibrium expression. It is important to recognize a buffer when you see one so you can use this alternative approach to calculating its pH!

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution and is given as

$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{\textcolor{green}{A^-}}]}{\mathrm{[\textcolor{red}{HA}]}}$

where pKa is the strength of the acid.

You can also determine the pOH of a buffer solution using

$\mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{\textcolor{red}{HB^+}}]}{\mathrm{[\textcolor{green}{B}]}}$

where pKb is the strength of the base.

Derivation of Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation can be derived from the equilibrium expression for the acid-ionization reaction.

\begin{align*} \dfrac{[\mathrm{H}^+][\mathrm{A^-}]}{[\mathrm{HA}]} &= K_{\mathrm{a}} \\ [\mathrm{H}^+] &= \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} K_{\mathrm{a}} \\ -\log [\mathrm{H}^+] &= -\left [ \log \left ( \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} K_{\mathrm{a}} \right ) \right ] \\ \mathrm{pH} &= -\left [ \log\left ( \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} \right ) + \log K_{\mathrm{a}} \right ] \\ &= -\log\left ( \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} \right ) - \log K_{\mathrm{a}} \\ &= -\log\left ( \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} \right ) + \mathrm{p}K_{\mathrm{a}} \\ \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\left ( \dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right ) \end{align*}

# How a buffer resists a change in pH

Part of this presentation was adopted from James Gaidis (https://chemistry.stackexchange.com/questions/81192).

## Weak acid solution

An solution made from a weak acid is represented by the following chemical reaction

$\mathrm{HA} (aq) + \mathrm{H_2O} (l) \rightleftharpoons \mathrm{H_3O^+} (aq) + \mathrm{A^-} (aq)$

and has the following equilibrium expression

$K_{\mathrm{a}} = \dfrac{[\mathrm{A^-}][\mathrm{H_3O^+}]}{[\mathrm{HA}]}$

where HA represents the acid and A represents the conjugate base.

Let us consider a case where the weak acid has a $$K_{\mathrm{a}} = 1 \times 10^{-6}$$ such that

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{[\mathrm{A^-}][\mathrm{H_3O^+}]}{[\mathrm{HA}]}$

If the solution is a 1 M solution of HA then concentration of A is 0.001 M such that

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(0.001)[\mathrm{H_3O^+}]}{1.00} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 1\times 10^{-3} \mathrm{~and~pH} = 3.0$

The concentration of A is three orders of magnitude smaller than HA. Additionally, a small amount of H3O+ is generated from the weak acid and drops the pH of pure water from 7 to 3 for the solution. This small increase in H3O+ leads to a large change in pH.

### Add 0.1 mol strong base

A strong base, when added to the solution, will react with the weak acid that is present in the system Adding 0.1 mol strong base to the solution causes the pH to increase. (Note: values from this point forward are approximate.)

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(0.1)[\mathrm{H_3O^+}]}{0.9} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 9.1\times 10^{-6} \mathrm{~and~pH} = 5.04$

A pH change of 2 is observed for this solution.

### Add 0.0005 mol strong acid

A strong acid, when added to the solution, will react with the conjugate base (A) that is present in the system Adding 0.0005 mol strong acid to the solution causes the pH to decrease. Note: The amount of strong acid added must be tiny since [A] is already tiny (i.e. 0.001 M).

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(5.0\times 10^{-4})[\mathrm{H_3O^+}]}{1.0005} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 2.0\times 10^{-3} \mathrm{~and~pH} = 2.70$

Adding an extremely small amount of strong acid gives rise to a pH change of 0.30.

## Weak acid buffer

Now consider a 1 L buffer solution with 0.5 M concentrations of HA and A (the conjugate base introduced from a salt) such that

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(0.5)[\mathrm{H_3O^+}]}{0.5} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 1.0\times 10^{-6} \mathrm{~and~pH} = 6$

### Add 0.1 mol strong base

A strong base, when added to the buffer, will react with the weak acid that is present in the system. If 0.1 mol of a strong base is added, HA is consumed and A increases.

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(0.6)[\mathrm{H_3O^+}]}{0.4} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 0.66\times 10^{-6} \mathrm{~and~pH} = 6.17$

A pH change of 0.17 is observed for this buffer (compared to a change of 2 for the non-buffer solution discussed above).

### Add 0.1 mol strong acid

A strong acid, when added to the buffer, will react with the conjugate base that is present in the system. If 0.1 mol of a strong acid is added, HA is produced and A is consumed.

$K_{\mathrm{a}} = 1\times 10^{-6} = \dfrac{(0.4)[\mathrm{H_3O^+}]}{0.6} \;\; \therefore \;\; \mathrm{[H_3O^+]} = 1.5\times 10^{-6} \mathrm{~and~pH} = 5.82$

Adding a relatively large amount of strong acid (relatively large compared to the amount we added to the solution above) causes a pH change of 0.18 for this buffer. This pH change is smaller than the pH change seen for the solution (where the pH change was 0.30) despite the fact that here we added an amount of acid that was three orders of magnitude greater!

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