According to Hasselbach-Henderson equation:
![pH=pK_{a}+log\frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Here,
is concentration of conjugate base and [HA] is concentration of acid.
In the given problem, conjugate base is
and acid is
thus, Hasselbach-Henderson equation will be as follows:
...... (1)
(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,
![\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOK%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%3D2)
Also, 
Putting the values in equation (1),

Therefore, pH of solution is 5.06.
(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,
![\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOK%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%3D%5Cfrac%7B1%7D%7B3%7D)
Also, 
Putting the values in equation (1),

Therefore, pH of solution is 4.28.
(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,
![\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOK%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%3D%5Cfrac%7B5%7D%7B1%7D)
Also, 
Putting the values in equation (1),

Therefore, pH of solution is 5.45.
(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,
![\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOK%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%3D1)
Also, 
Putting the values in equation (1),

Therefore, pH of solution is 4.76.
(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,
![\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOK%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%3D%5Cfrac%7B1%7D%7B10%7D)
Also, 
Putting the values in equation (1),

Therefore, pH of solution is 3.76.