Question

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).

Answer 1

when propionic acid is [HA] = 0.09 m 

potassium propionate is [A-] = 0.065 m 

and we have Ka = 1.3 x 10^-5

So we can use H-H equation:

PH = Pka + ㏒ [A-/HA]

when Pka = -㏒ Ka 

∴Pka = -㏒ (1.3 x 10^-5)
         =  4.89

by substitution:

∴PH = 4.89 + ㏒[0.065/ 0.09]

        = 4.75 

Answer 2

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

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