Question

"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"

Answer 1

Dissociation of butyric acid can be represented as:

HC₄H₇O₄------> H⁺ + C₄H₇O₄⁻

So its dissociation constant can be calculated as:

Ka={[H⁺][C₄H₇O₄⁻ ]}/[HC₄H₇O₄]

As pH = -log[H⁺]

2.71 = -log[H⁺]

[H⁺]= 0.0019

[H⁺]=[C₄H₇O₄⁻ ]=0.0019

As [HC₄H₇O₄]= 0.25M

So Ka={[H⁺][C₄H₇O₄⁻ ]}/[HC₄H₇O₄]

=(0.0019×0.0019)/0.25

=0.000014

Answer 2

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$[/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .