Respuesta :
Explanation:
The given data is as follows.
Moles of chloroacetic acid = 0.08 mol
Moles of sodium chloroacetate = 0.04 mol
So, first calculate the molarity of chloroacetic acid and sodium chloroacetate as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
Molarity of chloroacetic acid = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.08 mol}{1.00 L}[/tex]
= 0.08 M
Molarity of sodium chloroacetate = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.04 mol}{1.00 L}[/tex]
= 0.04 M
Now, according to the Henderson-Hasselbalch equation,
pH = [tex]pK_{a} + log(\frac{\text{base}}{\text{acid}})[/tex]
= [tex]pK_{a} + log{\frac{[salt]}{[acid]}}[/tex]
= [tex]2.865 + log \frac{0.04}{0.08}[/tex]
= 2.564
Thus, we can conclude that pH of given solution is 2.546.
