Respuesta :
Explanation:
It is given that the total volume is (10 mL + 60 mL) = 70 mL.
Also, it is known that [tex]M_{1}V_{1}[/tex] = [tex]M_{2}V_{2}[/tex]
Where, [tex]V_{1}[/tex] = total volume
[tex]V_{2}[/tex] = initial volume
Therefore, new concentration of [tex]CH_{3}COOH[/tex] = [tex]\frac{M_{2}V_{2}}{V_{1}}[/tex]
= [tex]\frac{60 \times 0.5}{70}[/tex]
= 0.43 M
New concentration of NaOH = [tex]\frac{M_{2}V_{2}}{V_{1}}[/tex]
= [tex]\frac{10 \times 1.0}{70}[/tex]
= 0.14 M
So, the given reaction will be as follows.
[tex]CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O[/tex]
Initial: 0.43 0.14 0
Change: -0.14 -0.14 0.14
Equilibrium: 0.29 0 0.14
As it is known that value of [tex]pK_{a}[/tex] = 4.74
Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.
pH = [tex]pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}[/tex]
= [tex]4.74 + log \frac{0.14}{0.29}[/tex]
= 4.74 + (-0.316)
= 4.42
Therefore, we can conclude that the pH of given reaction is 4.42.
