Calculate the ph of a solution that is 0.065 m in potassium propionate (c2h5cook or kc3h5o2) and 0.090 m in propionic acid (c2h5cooh or hc3h5o2).

The pH of the solution with 0.065 m in potassium and 0.090 m in propionic acid is : 4.75

Given data :

Potassium propionate [ A⁻ ] : 0.065 m

Propionic acid [ HA ] : 0.090 m

Ka = 1.3 * 10⁻⁵

To determine the pH of the solution we will apply the H-H equation

pH = Pka + log [ A / HA ] ( 1 )

where : pKa = -log Ka = 4.89

Insert value into equation ( 1 )

pH = 4.89 + log [ 0.065 / 0.090 ]

     = 4.75

Therefore the pH of the solution is 4.75

Learn more about pH of solution : link

6.4 cm (max) - 25 cm (min)