Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Step-by-step explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![pH=pKa+log(([Base])/([Acid]) )](https://img.qammunity.org/2021/formulas/chemistry/college/km74gaxjix359lr99r0my3yacfx5dsp8u5.png)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a.
![([Base])/([Acid])=(0.497M)/(0.365M)=1.36](https://img.qammunity.org/2021/formulas/chemistry/college/tgkjharx0hqjwyjvdor9ysb5u0o0lat2os.png)
b.
![([Base])/([Acid])=(0.217M)/(0.521M)=0.417](https://img.qammunity.org/2021/formulas/chemistry/college/odgxgjldah2vqdhc5s0bvhbrgv5d3g06sy.png)
c.
![([Base])/([Acid])=(0.713M)/(0.821M)=0.868](https://img.qammunity.org/2021/formulas/chemistry/college/qfkz9227y7khrxkwu31pxv56sl3bwnj3wr.png)
d.
![([Base])/([Acid])=(0.116M)/(0.121M)=0.959](https://img.qammunity.org/2021/formulas/chemistry/college/htpe8ghgra9ydc9756lmbnbgb0smbxzo6t.png)
Therefore, the d. solution has the best buffering capacity.
Regards.