Buffer solutions and calculations

In a nutshell

Buffer solutions are solutions which are able to minimise changes in pH when small amounts of acid or alkali are added.

What is a buffer solution?

A buffer is a solution which is able to minimise changes in pH when small amounts of acid or alkali are added. Buffers are made up of a weak acid or base and the conjugate base or acid. Acidic buffers have a pH of less than $$7$$, while basic buffers have a pH greater than $$7.$$​​

Acidic buffer solutions

Acidic buffer solutions have a pH of less than $$7$$ and they contain a weak acid and a conjugate base. There are two ways a buffer can be made using a weak acid. Either via the addition of a salt of the conjugate base or mixed with a strong base:

Salt of the conjugate base:

​$$Acid\ dissociation:\ H_3C-COOH \leftrightharpoons\ H_3C-COO^-\ +\ H^+ \\Conjugate\ base:\ H_3C-COO^-Na^+ \rightarrow\ H_3C-COO^-\ +\ Na^+$$​​

The salt fully dissociates in solution whilst the acid only partially dissociates.

Using a strong base:

​$$Acid\ and\ strong\ base:\ H_3C-COOH\ +\ OH^-\ \rightarrow\ H_3C-COO^-\ +\ H_2O \\Weak\ acid\ dissociation:\ H_3C-COOH\ \leftrightharpoons\ H_3C-COO^-\ +\ H^+$$​​

The weak acid is in excess, therefore there is unreacted acid left over in solution.

Acid conjugate base equilibrium

It is important to understand how the acid conjugate base equilibrium will be affected when small amounts of acid and base/alkali are added.

​$$H_3C-COOH\ \leftrightharpoons\ H_3C-COO^-\ +\ H^+$$​​

If an acid is added to the solution, it will favour the formation of the acid:

​$$H_3C-COOH\ \leftarrow\ H_3C-COO^-\ +\ H^+$$​​

Acid causes an increase in the concentration of $$H^+$$​ ions in solution. This leads to the formation of the weak acid (ethanoic acid in this case), which results in a shift to the left-hand side of the equilibrium as more hydrogen ions are released.

If a base is added to the solution, it will favour the formation of protons and conjugate base:

​$$H_3C-COOH\ \rightarrow\ H_3C-COO^-\ +\ H^+$$​​

The base reacts with the hydrogen ions in solution which decreases the concentration of $$H^+$$​ ions in solution. This leads to the dissociation of the weak acid (ethanoic acid in this case), which results in a shift to the right-hand side of the equilibrium as more hydrogen ions are released.

Alkaline buffer solutions

​Alkaline buffer solutions have a pH of greater than $$7$$​ and they contain a weak base and a conjugate acid. These work similarly to that of acidic buffers.

Consider a solution of ammonia $$(NH_3)$$​ and ammonium chloride $$(NH_4Cl)$$:

​$$NH_4Cl\ \leftrightharpoons\ NH_4^+\ +\ Cl^-$$ 

$$NH_3\ +\ H^+\ \leftrightharpoons\ NH_4^+$$ 

If a small amount of acid is added to the solution, the formation of ammonium will be favoured:

​$$NH_3\ +\ H^+\ \rightarrow\ NH_4^+$$​​

The formation of ammonium is favoured to remove hydrogen ions from the solution as small amounts of $$H^+$$ions are added. Therefore the pH can be kept near constant as the acid is added.

If a small amount of base is added to the solution, the formation of ammonia and protons will be favoured:

​$$NH_3\ +\ H^+\ \leftarrow\ NH_4^+$$​​

Buffers in human blood

Blood needs to be kept around a pH range of $$7.35\ -\ 7.45$$. The pH is controlled using a carbonic acid and hydrogen carbonate buffer:

  $$H_2CO_3\ \rightleftharpoons\ H^+\ +\ HCO_3^-$$​

Carbonic acid is formed when carbon dioxide mixes with water in the bloodstream $$(H_2O\ +\ CO_2\ \rightarrow\ H_2CO_3)$$. Since carbon dioxide is produced when respiration occurs, it causes an increase in the concentration of carbonic acid in the blood. To keep the pH constant in the blood, the equilibrium will shift to produce carbonic acid:

 $$H_2CO_3\ \leftarrow\ H^+\ +\ HCO_3^-$$​​

To increase the acidity of the blood, the equilibrium will shift to release more protons into the blood:

​$$H_2CO_3\ \rightarrow\ H^+\ +\ HCO_3^-$$​​

How to calculate the pH of a buffer

Calculating the pH of a buffer is not too difficult as long as the $$K_a$$ of the weak acid is known and the concentrations of the salt and the acid. To calculate the pH these two equations are used:

​$$[H^+]\ =\ K_a\ \times\ \frac{[HA]}{[A^-]}$$​​

​$$pH\ =\ -log_{10}([H^+])$$ 

Example

A buffer solution contains $$0.5\ mol\ dm^{-3}$$ of benzoic acid $$(C_6H_5COOH)$$​ and $$0.8 \ mol\ dm^{-3}$$ of sodium benzoate $$(C_6H_5COO^-Na^+)$$​. The $$K_a$$ of benzoic acid is $$6.3\ \times\ 10^{-5} \ mol \ dm^{-3}$$. Determine the pH of the buffer solution.

Firstly, the concentration of hydrogen ions in solution needs to be calculated:

$$[H^+]\ =\ K_a\ \times\ \frac{[HA]}{[A^-]}\\$$ 

$$[H^+]\ =\ 6.3\ \times\ 10^{-5}\ \times\ \frac{0.5}{0.8}\ =\ 3.9375\ \times\ 10^{-5}\ mol\ dm^{-3} = 3.94 \times\ 10^{-5}\ mol\ dm^{-3} (2dp)$$ 

The pH can be calculated now that the concentration of hydrogen ions is known.

$$pH\ =\ -log_{10}([H^+])$$ 

$$pH\ =\ -log_{10}(3.94\ \times 10^{-5})\ =\ \underline{4.40}$$ 

The pH of the solution is $$\underline{4.40}$$.​

How to calculate the concentration of an acid

If a buffer has to be a specific pH the concentration required for the acid can be calculated from the $$K_a$$ of the acid used and the concentration of conjugate base in solution $$([A^-])$$. The Henderson-Hasselbalch equation is used to calculate the concentration of the acid.

​$$pH\ =\ pK_a\ +\ log_{10}(\frac{[A^-]}{[HA]}$$​

Example

A buffer with a pH of $$4.1$$ is made from methanoic acid $$(HCOOH)$$ and potassium methanoate, the conjugate base $$(HCOO^-K^+)$$. $$1.6 \ mol\ dm^{-3}$$​ of conjugate base is used and the $$K_a$$ of the acid is $$1.8\ \times\ 10^{-4}\ mol\ dm^{-3}$$. Determine, the concentration of methanoic acid required to form the buffer.

Firstly, convert the $$K_a$$ of the acid to $$pK_a$$, this is so that it can be used in the Henderson-Hasselbalch equation.

$$pK_a\ =\ -log(K_a) \\pK_a\ =\ -log(1.8\ \times\ 10^{-4})\ =\ 3.74$$​

Next, the $$pK_a$$ and pH can be inputted into the Henderson-Hasselbalch equation and re-arranged to give the ratio of $$([A^-]:[HA])$$.​

$$pH\ =\ pK_a\ +\ log_{10}(\frac{[A^-]}{[HA]}) \\log_{10}(\frac{[A^-]}{[HA]})\ =\ pH\ -\ pK_a \\$$ 

The log rule states that:

$$10^{log_{10}x}\ =\ x$$, in this case $$x$$ is $$(pH\ -\ pK_a)$$.

So to get the ratio of $$([A^-]:[HA])$$:

$$\frac{[A^-]}{[HA]}\ =\ 10^{pH\ -\ pK_a}$$ 

This can be further re-arranged to give the concentration of the acid:

$$[HA]\ =\ \frac{[A^-]}{10^{pH\ -\ pK_a}}$$ 

Inputting the values from the equation gives:

$$[HCOOH]\ =\ \frac{1.6\ mol\ dm^{-3}}{10^{4.41\ -\ 3.74}} \\$$ 

$$[HCOOH]\ =\ \frac{1.6\ mol\ dm^{-3}}{2.29}\ =\ \underline{0.70\ mol\ dm^{-3}}$$ 

To achieve the required pH of $$4.1$$, $$\underline{0.70\ mol\ dm^{-3}}$$ of methanoic acid is required.​

Note: Using the unrounded values in your working out will yield a value of $$0.71 \ mol \ dm^{-3}\ (2 \ dp)$$.