Calculate the pH of a 0.002M acetic acid solution if it is 2.3% ionised at this dilution, ka=1.8×10^-5.

3.11 is the pH of the solution

Explanation:

The ionised form of the acetic acid is the acetate ion. To solve this problem we need to solve H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer

pKa = -log Ka  = 4.74

[A⁻] = Concentration of acetate ion = 0.002M * 2.3% = 4.6x10⁻⁵M

[HA] = Concentration of acetic acid = 0.002M *(100-2.3)% = 0.001954M

Replacing:

pH = 4.74 + log [4.6x10⁻⁵M] / [0.001954M]

pH = 3.11 is the pH of the solution

particle   p is an electron and particle q is a neutron