# pH Calculation - buffer solutions

A buffer solution is a solution which is not significantly affected by changes in pH due to dilutions or small additions of strong acids or bases.

There are two types of buffer solution:

• those prepared by mixing a weak acid and a salt of its conjugate base (eg: CH3COO- / CH3COOH)
• those prepared by mixing a weak base and a salt of its conjugate acid (eg NH3/NH4+)

In order to make a buffer that is actually able to buffer the pH  significant amounts must be prepared by mixing up the two species that constitute it.

• how does a buffer solution work

Take for example the buffer solution made by acetic acid / acetate. After dissolution in water the respective balances are:
- acetic acid $\bg_white&space;CH_3COOH&space;+&space;H_2O&space;\rightleftharpoons&space;CH_3COO^-&space;+&space;H_3O^+$

- acetate

$\bg_white&space;\bg_white&space;CH_3COO^-&space;+&space;H_2O&space;\rightleftharpoons&space;CH_3COOH&space;+&space;OH^-$

As you can see, $\bg_white&space;CH_3COO^-$ act as common ion in the dissociation equilibrium of acetic acid, pushing the balance towards the left, while $\bg_white&space;CH_3COOH$ figures in the balance of acetate hydrolysis, and also in this case it pushes the reaction to the left.

So, we can assume that when we prepare a buffer, the concentrations of the two species once the equilibrium is reached are equal to the respective initial concentrations, because of the common ion effect.

- But how do they resist the addition of strong acid?

Putting a strong acid in solution means putting H+ ions in solution. These will be neutralized by CH3COO- , given that it is a base. The effect of the acid is not marked,  and considering this balance:

$\bg_white&space;CH_3COOH&space;+&space;H_2O&space;\rightleftharpoons&space;CH_3COO^-&space;+&space;H_3O^+$

It will move to the right.

-and the addition of a strong base?

It takes place exactly the opposite. The hydroxyl ion OH- released from the base in solution react quantitatively with CH3COOH. Even in this case, in response to this reaction, the equilibrium:

$\bg_white&space;\bg_white&space;CH_3COO^-&space;+&space;H_2O&space;\rightleftharpoons&space;CH_3COOH&space;+&space;OH^-$

will move to the right.

-and what will it happen diluting the solution?

Dilution has no effect on pH of a buffer solution since it decreases at the same time the concentration of the acid and that the salt (the conjugate base). The ratio remains constant. This property will be better understood observing the formula for the calculation of pH.

A weak base - conjugate acid buffer will work in a similar way; the weak base will be able to counteract small additions of strong acid and its conjugate acid will be able to counteract small additions of strong bases.

pH calculation

• weak acid + conjugate base buffer

The concentration of hydrogen ions (H3O+) of a buffer solution consisting of a weak acid and the salt of its conjugate base is calculated as:

$\bg_white&space;[H_3O^+]&space;=&space;K_a&space;\cdot&space;\frac{C_a&space;}{C_s}$

Where Ka is the acid dissociation constant, Ca is the initial concentration of weak acid and Cs is the concentration of the salt. From this equation you can derive the formula for a straight pH calculation:

$\bg_white&space;pH&space;=&space;pK_a&space;+&space;log\left&space;(&space;\frac{C_s}{C_a}&space;\right&space;)$

Since the volume of solution is the same for both the acid and the salt the hydrogen ion concentration can also be derived as:

$\bg_white&space;[H_3O^+]&space;=&space;K_a&space;\cdot&space;\frac{n_a}{n_s}$

where obviously na are the moles of acid and ns are the moles of salt.

•  weak base +conjugate acid buffer

The concentration of hydroxyl ion (OH -) of a buffer solution made of a weak base and the salt of its conjugate base can be calculated as:

$\bg_white&space;\bg_white&space;[OH^-]&space;=&space;K_b&space;\cdot&space;\frac{C_b}{C_s}$

where Cb is the concentration of weak base and Cs is the concentration of the salt of its conjugate acid. In the same way we can derive the formula for calculating the pOH:

$\bg_white&space;pOH&space;=&space;pK_b&space;\cdot&space;\frac{C_s}{C_b}$

Also in this case it is possible to simplify the calculation of the concentration of $\bg_white&space;OH^-$ , given that the base and its salt are obviously dissolved in the same volume of solution:

$\bg_white&space;[OH^-&space;]&space;=&space;K_b&space;\cdot\frac{&space;n_b}{n_s}$

where nb are the basic moles and ns are the moles of salt.