Question

Butyric acid is responsible for the odor in rancid butter. A solution of 0.25 m butyric acid has a ph of 2.71. What is the ka for the acid?

Answer 1

Ka = 1.5 -10^-5

Step-by-step explanation:

Step 1: Data given

Molarity of the solution = 0.25 M

pH = 2.71

Step 2 The equation

C3H7COOH + NaOH ⇆ C3H7COONa + H2O

Step 3: Calculate pKa

pH = (pKa-log[acid])/2

2pH = pKa -log[acid]

pKa = 2pH +log[acid]

⇒with pH = 2.71

⇒with log[acid] = -0.60

pKa=2*2.71 -0.60

pKa = 5.42 - 0.60

pKa = 4.82

Step 4: Calculate Ka

pKa = -log(Ka)

Ka = 10^-4.82

Ka = 1.5 -10^-5