Answer:
Ka3 for the triprotic acid is 7.69*10^-11
Explanation:
Step 1: Data given
Ka1 = 0.0053
Ka2 = 1.5 * 10^-7
pH at the second equivalence point = 8.469
Step 2: Calculate Ka3
pKa = -log (Ka2) = 6.824
The pH at the second equivalence point (8.469) will be the average of pKa2 and pKa3. So,
8.469 = (6.824 + pKa3) / 2
pKa3 = 10.114
Ka3 = 10^-10.114 = 7.69*10^-11
Ka3 for the triprotic acid is 7.69*10^-11
Answer:
The answers are in the explanation
Explanation:
- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.
- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.
- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.
- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.
I hope it helps!
Answer: See image attached!
Good luck with your future exams!
Are you speaking of a density gradient, in which a more concentrated solution moves below a less concentrated solution?
In that case, the more concentrated solution has the greater density, and it will gradually sink below the less concentrated solution.
In the same way, a stone will sink in water, which is less dense than the stone.
Answer:B
Explanation: Because you know how the is potential energy and then there is kinetic energy yeah those have to do with movement like a roller coaster
