pH calculation - Hydrolysis of salts

A salt is a compound formed by ions, electrically neutral, the product of an acid-base reaction. The salts are strong electrolytes, and placed in water (if soluble) will completely dissociate into their constituent ions, ensuring the solution a certain electrical conductivity. This is regardless of whether they are made from weak or strong acids and weak or strong bases.

In general, the salts may be formed from:

1 - strong acid and strong base

2 - strong acid and weak base

3 - strong base and weak acid

4 - weak acid and weak base

The pH imparted by a salt to a solution is due to the interaction of the ions constituting the salt with H2O.

Let's see how to calculate the pH for aqueous solutions of these four different kind of salts.

1 - salt of strong acid and strong base

Let's take the simplest example absolute; sodium chloride, NaCl. The constituent ions are Na+ and Cl-. The acid-base theory of Bronsted-Lowry has taught us that the more strong the base the weaker it will be its conjugate acid and vice versa, the more strong the acid the weaker will be its conjugate base.

In this case, we have Na+, which is the conjugate acid of NaOH, and Cl-, which is the conjugate base of HCl. NaOH is a strong base, and in the same way HCl is a strong acid. It follows that the Na+ acid behavior and Cl- basic behavior are totally irrelevant.

If none of the two ions interacts with H2O then we will have no pH changes.

It follows that the pH of an aqueous solution of NaCl is neutral, 7.

The salts derived from strong acid and strong base leave unchanged the pH of a solution.

2 - salt of weak acid and strong base

Take for example the sodium acetate, CH3COONa. And 'it is including respectively CH3COO- and Na+. Na+ is the conjugate acid of NaOH, and as we saw in the previous example, its contribution to the pH is null. CH3COO- instead it is the conjugate base of CH3COOH, a weak acid. As the conjugate base of a weak acid, it shows a certain affinity towards the proton, and it behaves then as a base:

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

The acid-base reaction of CH3COO- with water is called "hydrolysis." In this case we are speaking of basic hydrolysis, because the reaction generates an excess of OH.Since the reaction generates this excess, the salt of a weak acid and a strong base in water give a pH> 7, ​​therefore alkaline. For the calculation of pH, we can consider the solution of this salt exactly as the solution of just a weak base (in our case CH3COO-). We have seen that the formula for calculating $\bg_white&space;[OH^-]$ generated resulting from a weak base is:

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

The Kb of the weak base CH3COO- can be obtained from Ka of acetic acid as it follows:

$\bg_white&space;K_w&space;=K_b&space;\cdot&space;K_a$$\bg_white&space;K_b&space;=&space;\frac{K_w}{K_a}$

The base concentration $\bg_white&space;C_b$ in this case is the concentration $\bg_white&space;C_s$ of our salt.

From this considerations the formula for calculating the $\bg_white&space;[OH^-]$ a salt-strong base weak acid solution:

$\bg_white&space;\bg_white&space;\bg_white&space;[OH^-]&space;=&space;\sqrt{\frac{K_w}{K_a}\cdot&space;C_s}$

from which you can derive the pH every time you have this specific circumstances.

3 - salt of weak base and strong acid

Take for example the ammonium chloride, NH4Cl. Its constituent ions are NH4+ and Cl.NH4+ is the conjugate acid of NH3, ammonia, a weak base. It will therefore have a tendency, altough small, to cede that acid hydrogen. Cl- instead is the conjugate base of HCl, and as we have seen in the examples above its contribution to the pH is practically zero, and then we can overlook it. We can simply consider a solution of this salt as an aqueous solution of the weak acid NH4+ (which will then give an acid pH)

$\bg_white&space;NH_4^+&space;+&space;H_2O&space;\rightleftharpoons&space;NH_3&space;+&space;H_3O^+$

The approximate formula to calculate the pH of a weak acid is the following:

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

The Ka of the ammonium ion can be easily obtained from the Kb of ammonia. The concentration of acid in the formula, Ca , is therefore in this case the concentration Cs of our salt.

$\bg_white&space;\bg_white&space;\bg_white&space;\bg_white&space;[H_3O^+]&space;=&space;\sqrt{\frac{K_w}{K_b}\cdot&space;C_s}$

4 - salt of weak acid and weak base

Let's take for example ammonium acetate, CH3COONH4. The constituent ions are CH3COO- and NH4+. Both of these ions are able to give hydrolysis (as we have seen in the previous examples); CH3COO- gives basic hydrolysis, while acid hydrolysis is given by NH4.The pH of a salt composed of weak acid and weak base will depend on the strength relationship of the acidic and basic components of the salt. If the base strength and the acid strength are equal the result is a neutral solution. And that's right the case for example of CH3COONH4 .

Ka NH4 =  Kb CH3COO-

If base strength is prevailing on acid strength the salt will give the solution an alkaline pH. And that's the case for example of NH4CN.

Kb CN-  > Ka NH4+

If the prevailing acid strength on the base strength of course the result will be an acid pH.

In any case if we want to exactly calculate the pH of a salt formed by weak acid and weak base (also what we called a neutral solution if we wanted to be accurate, has to be calculated like this because Ka and Kb are always at least slightly different), there is a very simple formula:

$\bg_white&space;[H_3O^+]&space;=&space;\sqrt{K_w&space;\cdot&space;\frac{K_a}{K_b}}$

In which:

- Ka is the constant of the conjugate acid of the weak base which forms the salt. For example, in the case of CH3COONH4 would be the Ka of CH3COOH.

- Kb is the constant of the conjugate base of the weak acid that constitutes the salt. For example, in the case of CH3COONH4 would be the Kb of NH3.

In summary, when:

• Ka > Kb the resulting solution is acidic.
• Kb > Ka the resulting solution is alkaline.
• Ka = Kb the resulting solution is neutral.