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

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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