Answer:
A: pH = 3.74
B: pH = 5.74
C: pH = 4.74
Step-by-step explanation:
According to the Henderson-Hasselbach equation for buffers, we know that
. Firstly, let's find the
value:
![pK_a = -log(K_a) = -log(1.8\cdot 10^-5) = 4.74](https://img.qammunity.org/2020/formulas/chemistry/high-school/rgki9xqrb9wuxko6mxmw0aq97t6ied10rp.png)
Now, for buffer A:
![([A^-])/([HA])=([CH_3COO^-])/([CH_3COOH])=(1)/(10)](https://img.qammunity.org/2020/formulas/chemistry/high-school/xq3gyfm0urpp2myhsulczp6l3swapy07ch.png)
This means:
![pH = 4.74 + log((1)/(10)) = 3.74](https://img.qammunity.org/2020/formulas/chemistry/high-school/ycp6jsy8kl2fwk4ja1uznj9euniq6xcnms.png)
Similarly, for buffer B:
![([A^-])/([HA])=([CH_3COO^-])/([CH_3COOH])=(10)/(1)](https://img.qammunity.org/2020/formulas/chemistry/high-school/bkhykm9pnheuigzkto2928otc8ztpr8k2p.png)
This means:
![pH = 4.74 + log((10)/(1)) = 5.74](https://img.qammunity.org/2020/formulas/chemistry/high-school/7clpfobo7x3qwtcswni7x6kz6j04e2z4qr.png)
Similarly, for buffer C:
![([A^-])/([HA])=([CH_3COO^-])/([CH_3COOH])=(1)/(1)](https://img.qammunity.org/2020/formulas/chemistry/high-school/obw7tts4yaq6rv51bm56xlrn4mg48r0d3k.png)
This means:
![pH = 4.74 + log((1)/(1)) = 4.74](https://img.qammunity.org/2020/formulas/chemistry/high-school/gybofmimpi0vvxrwsdikqhb13bl4dplpww.png)