# pH Calculation - Weak Acid

Weak acids are those acids which in solution does not dissociate completely. As acids (Bronsted-Lowry / Arrhenius theories), in solution they donate a proton to H2O (A is a generic counterion, that does not give hydrolysis):

$HA&space;+&space;H_2O&space;\rightleftharpoons&space;H_3O^+&space;+&space;A^-$

The acid dissociation constant is given by:

$Ka&space;=&space;\frac{[H_3O^+][A^-]}{[HA]}$

Take for example acetic acid, a very common weak acid:

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

In presence of water the acetic acid will dissociate into his constituents ions $CH_3COO^-$  and $H_3O^+$ , while in part will remain undissociated as:$CH_3COOH$

Then we will have to find out the balance achieved in solution between the species $CH_3COO^-$ , $H_3O^+$  and  $CH_3COOH$ . To calculate the pH we just need the equation of Ka:

$Ka&space;=\frac{&space;[CH_3COO^-][H_3O^+]}{[CH_3COOH]}&space;=&space;1,8&space;\cdot&space;10^{-5}$

We must therefore find $[H_3O^+]$ as pH is equal to -log [H3O+].

We call Ca is the initial molar concentration of acid, before the dissociation in water. We denote instead with "X" the amount of acid that dissociates.

If X mol/L of acetic acid dissociate, we will get X mol/L of H3O+ and X mol/L CH3COO-.

So we will have X = [H3O+] = [CH3COO-]

The concentration of CH3COOH that remains undissociated will be (Ca - X) mol/L

Substituting in the expression of the Ka:

$Ka&space;=\frac{&space;[x][x]}{C_a&space;-&space;[x]}&space;=\frac{&space;[x^2]}{C_a&space;-&space;[x]}&space;=&space;1,8\cdot10^{-5}$

This is a quadratic equation that we are perfectly able to solve. It can be however, further simplificated. Since we are dealing with a weak acid, we can suppose that for the most part it remains undissociated.

This means that $[CH_3COOH]&space;>>&space;[H_3O^+]$ we can say with good approximation that:

$C_a&space;-&space;x&space;=&space;C_a$

Our expression therefore becomes:

$Ka&space;=\frac{&space;[x^2]}{C_a}$

and given that X = [H3O+]

$Ka&space;=\frac{&space;[H_3O^+]^2}{C_a&space;}$ that, rearranged as a function of $[H_3O^+]$ becomes: $[H_3O^+]=&space;\sqrt{Ka\cdot&space;C_a}$

It can be derived that the final formula to calculated calculate the pH is:

$pH&space;=&space;\frac{1}{2}(pKa&space;-&space;logC_a)$

• pH calculation of a weak base

The pH of a weak base can be calculated similarly, by the same steps and the same approximations. Firstly we calculate [OH- ]:

$[OH^-]&space;=&space;\sqrt{Kb\cdot&space;Cb}$

And consequently the pOH:

$pOH&space;=&space;\frac{1}{2}\cdot&space;(pK_b&space;-&space;logC_b)$

The pH will be simply found as:

$pH&space;=&space;14&space;-&space;pOH$