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Answer:
The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.
Explanation:
When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.
Hence, option B is correct.
Answer:
Explanation:
Hello there!
In this case, according to the Henderson-Hasselbach equation, it is possible to write:
![pH=pKa+log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Next, since we are given the pH and the [A–]/[HA] ratio, we can solve for the pKa as shown below:
![pKa=pH-log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pKa%3DpH-log%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Now, we plug in the values to obtain:
Next, Ka is:
Best regards!
The bond between Potassium and Bromide is considered iconic, Because of how they lose electrons and become ions when they bond together.
11.....6 for Na, 1 for S and 4 for O
