Question

The odor of spoiled butter is due in part to butanoic acid, which results from the chemical breakdown of butter fat. A 0.100 M solution of butanoic acid is 1.23% ionized. Calculate the equilibrium constant for this weak monoprotic acid.

Answer 1

Ionization reaction :

CH₃CH₂CH₂COOH ​​+ H₂O <==> H₃O⁺   + CH₃CH₂CH₂COO^-

^{Equilibrium constant(kc) is written as :}

kc=[H₃O⁺ ][CH₃CH₂CH₂COO^-]/[CH₃CH₂CH₂COOH][H₂O]----(1)

Acid is 1.23% ionized so there is 98.77 unionized acid.

[H⁺ ] = [CH₃CH₂CH₂COO^-] = 0.0123(0.100M)=0.00123 M

[CH₃CH₂CH₂COOH]=0.9877(0.100M)=0.09877 M

[H₂O]= 1M

Putting all these values in equation (1)

kc= [H⁺ ][CH₃CH₂CH₂COO^-]/[CH₃CH₂CH₂COOH][H₂O]

kc =(0.00123M)(0.00123M)/(0.09877M)(1M)

kc = 0.0000015129/0.09877

kc = (1.5129 x 10^-6)/0.09877

kc = 0.153 x 10 ^-4

^{Equilibrium constant of weak monoprotic acid is} 0.153 x 10 ^-4