You have to prepare a pH 5.00 buffer, and you have the following 0.10 M solutions available: HCOOH (pka=3.74), HCOONa, CH3COOH (
1 answer:
Answer:
CH3COOH - CH3COONa since its pKa is closest to the required pH.
Explanation:
Hello there!
In this case, in agreement with the theory of buffers as solutions able to withstand severe pH changes due to the addition of acidic or basic substances, it is possible to set up a generation equilibrium expression for the acids herein given:
Which leads to the Henderson-Hasselbach equation:
Thus, since all the buffers have [A-]=[HA]=0.10M, the log part becomes 0 and therefore the best buffer will have the closest pKa to the required pH, which is CH3COOH - CH3COONa since its pKa is 4.74.
Best regards!
You might be interested in
Answer:
It will require<u> second round</u> of the cycle to release
Explanation:
<u>Reason behind the requirement of second round of the cycle to release </u> -:
The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.