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 value of Ka for butanoic acid.

The value of Ka for butanoic acid is 1.53 X 10⁻⁵

Explanation:

Given-

Initial concentration of butanoic acid = 0.1M

Ionization percentage = 1.23%

Dissociation of butanoic acid:

                          HC₄H₇O₂ ⇒ C₄H₇O₂⁻ + H⁺

Initial                        0.1                 0          0

Change                    -x                  x           x

Equilibrium            0.1 - x              x           x

The value of x can be calculated from the percent ionization.

percent ionization = |H⁺| X 100 / |HC₄H₇O₂|

1.23% = |H⁺| X 100% / 0.1M

|H⁺| = 0.1M X 1.23% / 100%

|H⁺| = 0.00123M

|H⁺| = |C₄H₇O₂⁻|

|C₄H₇O₂⁻| , x= 0.00123 M

Solving for the equilibrium concentration of HC₄H₇O₂

|HC₄H₇O₂| = 0.1 - x

|HC₄H₇O₂| = 0.1 - 0.00123 M

|HC₄H₇O₂| = 0.09877 M

Ka can be calculated as:

Ka = |C₄H₇O₂⁻| |H⁺| / |HC₄H₇O₂|

Ka = (0.00123) X (0.00123) / (0.09877)

Ka = 1.53 X 10⁻⁵

The value of Ka for butanoic acid is 1.53 X 10⁻⁵

0.4M NaOH

Explanation:

Molarity, M, is an unit of concentration widely used defined as the ratio between moles of solute (In this case, NaOH) and volume of solution in liters.

As the solution contains 2 moles of NaOH-Moles of solute- in 5L of solution, the molarity is:

2 moles NaOH /  5L =