Answer:
The maximum volume is 122.73 mL
Explanation:
125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.
pKa of acetic acid = 4.76
Using Henderson Hasselbalch equation.
pH =pKa + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
= 4.76 + [tex]log \frac{0.3600}{0.3600}[/tex]
= 4.76
Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.
When HCl is added , it reacts with CH₃COO⁻ to give CH₃COOH.
However, [CH₃COOH] increases and the log term results in a negative value.
Let assume, the new pH is less than 4.76
Let say 3.76; calculating the concentration when the pH is 3.76; we have
3.76 = 4.76 + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 4.76 - 3.76
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 1.00
[tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 0.1
Let number of moles of acid be x (i.e change in moles be x); &
moles of acetic acid and conjugate base present be = Molarity × Volume
= 0.360 M × 125 mL
= 45 mmol
replacing initial concentrations and change in the above expression; we have;
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1[/tex]
[tex]\frac{45-x}{45+x} =0.1[/tex]
0.1(45 + x) = 45 - x
4.5 + 0.1 x = 45 -x
0.1 x + x = 45 -4.5
1.1 x = 40.5
x = 40.5/1.1
x = 36.82
So, moles of acid added = 36.82 mmol
Molarity = 0.300 M
So, volume of acid = [tex]\frac{moles}{molarity }[/tex] = [tex]\frac{36.82 \ mmol}{0.300}[/tex]
= 122.73 mL
