Respuesta :
According to Hasselbach-Henderson equation:
[tex]pH=pK_{a}+log\frac{[A^{-}]}{[HA]}[/tex]
Here, [tex][A^{-}][/tex] is concentration of conjugate base and [HA] is concentration of acid.
In the given problem, conjugate base is [tex]CH_{3}COOK[/tex] and acid is [tex]CH_{3}COOH[/tex] thus, Hasselbach-Henderson equation will be as follows:
[tex]pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}[/tex]...... (1)
(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,
[tex]\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2[/tex]
Also, [tex]pK_{a}=4.76[/tex]
Putting the values in equation (1),
[tex]pH=4.76+log\frac{2}{1}=5.06[/tex]
Therefore, pH of solution is 5.06.
(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,
[tex]\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}[/tex]
Also, [tex]pK_{a}=4.76[/tex]
Putting the values in equation (1),
[tex]pH=4.76+log\frac{1}{3}=4.28[/tex]
Therefore, pH of solution is 4.28.
(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,
[tex]\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}[/tex]
Also, [tex]pK_{a}=4.76[/tex]
Putting the values in equation (1),
[tex]pH=4.76+log\frac{5}{1}=5.45[/tex]
Therefore, pH of solution is 5.45.
(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,
[tex]\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1[/tex]
Also, [tex]pK_{a}=4.76[/tex]
Putting the values in equation (1),
[tex]pH=4.76+log\frac{1}{1}=4.76[/tex]
Therefore, pH of solution is 4.76.
(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,
[tex]\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}[/tex]
Also, [tex]pK_{a}=4.76[/tex]
Putting the values in equation (1),
[tex]pH=4.76+log\frac{1}{10}=3.76[/tex]
Therefore, pH of solution is 3.76.
