To determine the volume of 0.2 M acetic acid needed to create a buffer solution with a pH of 4.94 , we'll use the Henderson-Hasselbalch equation, which is:
pH=pKa+log(‌| [‌ salt ‌] |
| [‌ Acid ‌] |
)Here:
square pH=4.94pKa of acetic acid
=4.76[Salt] is the concentration of sodium acetate
[Acid] is the concentration of acetic acid
Given the volume and concentration of sodium acetate:
Volume of sodium acetate,
V‌salt ‌=100‌mlConcentration of sodium acetate,
C‌salt ‌=0.4MFirst, let's find the ratio of the concentrations of salt to acid using the Henderson-Hasselbalch equation:
‌4.94=4.76+log(‌| [‌ salt ‌] |
| [‌ acid ‌] |
)‌4.94−4.76=log(‌| [‌ salt ‌] |
| [‌ acid ‌] |
)‌0.18=log(‌| [‌ salt ‌] |
| [‌ acid ‌] |
)To remove the logarithm, we raise 10 to the power of both sides:
‌100.18=‌| [‌ salt ‌] |
| [‌ acid ‌] |
‌1.51=‌| [‌ salt ‌] |
| [‌ acid ‌] |
Let's denote the volume of acetic acid to be added as
V‌acid ‌ in milliliters. The concentration of acetic acid is given as 0.2 M . The total volume of the buffer solution will be
100‌ml+V‌acid ‌. Therefore, the concentrations of salt and acid in the buffer are:
‌[‌ Salt ‌]=‌| 0.4M×100‌ml |
| 100‌ml+V‌sid ‌ |
‌[‌ Acid ‌]=‌| 0.2M×V‌scd ‌ |
| 100‌ml+V‌scid ‌ |
Now, substituting these into the ratio we found:
1.51=‌‌| 0.4M×100‌ml | | 100‌ml∣+V‌acid ‌ |
|
| 0.2M×V‌scid ‌ | | 100‌ml∣V‌acid ‌ |
|
The
100‌ml+V‌acid ‌ terms cancel out, so we have:
‌1.51=‌| 0.4×100 |
| 0.2×V‌ecid ‌ |
‌1.51=‌‌1.51×0.2×V‌acid ‌=40‌0.302×V‌acid ‌=40Finally, solving for
V‌acid ‌ :
V‌acid ‌=‌≈132.45‌ml However, this value does not match any of the given options exactly. Typically, slight approximations may occur due to rounding off values during intermediate steps. Considering the options, the closest value is:
Option A: 132.1 ml