Answer:- The Ka for the acid is
.
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}](https://tex.z-dn.net/?f=Ka%20%3D%20%5BH%5E%2B%5D%5BA%5E-%5D%5Cfrac%7B1%7D%7BHA%7D)
Where, Ka is the acid ionization constant. Let's plug in the values.
![Ka = \frac{X^2}{0.25-X}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7BX%5E2%7D%7B0.25-X%7D)
Let's calculate the value of X first using the equation:
[/tex]
on taking antilog ob above equation we get:
![[H^+]=10^-^p^H](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5Ep%5EH)
![[H^+]=10^-^2^.^7^1](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5E2%5E.%5E7%5E1)
= 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}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%280.00195%29%5E2%7D%7B0.25-0.00195%7D)
![Ka=1.53*10^-^5](https://tex.z-dn.net/?f=Ka%3D1.53%2A10%5E-%5E5)
So, the value of Ka for butyric acid is
.