# Buffer solutions

A buffer solution is a solution that does not undergo significant pH changes when adding small amounts of strong acids, strong bases or diluted.

The buffer solutions can be constituted by:

• an acid weak and its conjugate base ​​same as saying "weak and an acid salt thereof"
• a weak base and its conjugate acid → same as saying "a weak base and a salt"

The characteristics of a buffer system

To try to understand what the properties are due as interesting of a buffer system, we try to calculate the pH of a solution consisting of 0.75 M CH 3 COOH and CH 3 COONa 0.5M.

Consider the expression of the equilibrium constant of acetic acid:

$\bg_white&space;K_a&space;=&space;\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$

If the riarrangiamo as a function of $\bg_white&space;[H_3O^+]$

$\bg_white&space;\bg_white&space;[H_3O^+]&space;=&space;K_a&space;\cdot&space;\frac{[CH_3COOH]}{[CH_3COO^-]}$

It is obtained a report that can be used to calculate the pH of a solution that contains both the weak acid $\bg_white&space;CH_3COOH$ that its conjugate base $\bg_white&space;\bg_white&space;CH_3COO^-$ . But of course, $\bg_white&space;[CH_3COOH]$ is $\bg_white&space;\bg_white&space;[CH_3COO^-]$  are not the initial concentrations of the two species, but the equilibrium concentrations!

But you can make some interesting observations:

Consider the equilibrium dissociation of acetic acid:

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

We have an initial concentration C in acetic acid of 0.75 M. The dissociation of acetic acid, also under normal conditions, it is very modest (K a = 1.8 · 10 -5). This means that for the most part remains as undissociated CH 3 COOH. This is all the more true if we consider that there is already in solution the ion CH 3 COO - from CH 3 COONa, which is receding balance to the left.

then we can say that:

$\bg_white&space;\bg_white&space;[CH_3COOH]&space;\cong&space;C_a$

We can make very similar considerations $\bg_white&space;CH_3COO^-$ It concluded that:

$\bg_white&space;\bg_white&space;[CH_3COO^-]&space;\cong&space;C_s$

It 'important to understand that these two equilibria are concatenated, that must always be met simultaneously. The two species act like one o'clock in common ion balance of the other, pushing back their balances so that you can consider the respective equilibrium concentrations comparable to those initials.

Formula for the calculation of the pH of a weak acid buffer - conjugate base

We can then write:

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

from which:

which is the known equation of Henderson-Hasselbach.

Substituting the numerical values ​​of our example, we have:

$pH&space;=&space;4,57$

Which it is a more 'or less intermediate value between that of a solution only from our acetic acid (0.75 M) and only of our acetate (0.5 M).

buffering effect

We have said that a tampon, within certain limits, is able to not undergo significant changes in pH due to dilution or addition of small / strong basic acid.

We try to understand why by analyzing the relationships we have just written:

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

• Pursuing a dilution would decrease both the concentration of C in which the acid base conjugate concentration, C s. Their ratio remains approximately constant, without significant effects (less than large dilutions), on pH.
• To understand what happens as a result of a small add a strong base or acid, let's take a practical example. Suppose we add 1 ml of 0.1 M HCl. HCl sends the H + ions in solution, and the only species that can react with H + is CH 3 COO -, in a neutralization reaction.

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

the neutralization reaction can be considered quantitative. This means that 0.001 mol / L of CH 3 COO - are destroyed at the expense of 0.001 mol / L of CH 3 COOH that are formed.

Then we have to review the equilibrium concentrations of the two species:

$[CH_3COOH]_{eq}&space;=&space;0,75&space;+&space;0,001&space;=&space;0,751&space;M$

$[CH_3COO^-]&space;=&space;0,5&space;-&space;0,001&space;=&space;0,499&space;M$

Calculating the pH with these new values

$pH&space;=&space;pKa&space;+&space;log&space;\left&space;(\frac{C_s}{C_a}&space;\right&space;)&space;=&space;4,57$

This small addition of acid left the pH substantially unchanged. We would have achieved the same result by adding a small amount of strong base. For example, adding a small amount of NaOH:

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

A share of $CH_3COOH$ simultaneously with the formation it would disappear in the same amount of $CH_3COO^-$ . For the rest, the calculations are exactly the same that we have seen in the previous example, but with roles reversed.