An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined to be 6.5e-07. The pH at the second equivalence point of the titration is measured as 8.181. What is Ka3 for this triprotic acid
1 answer:
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
You might be interested in
Answer:
1000L
Explanation:
the 1 is a sig fig and since the 0 is between the 1 and 4 its also a significant number. to round them off you look at the 0,then look back at the 4 since its less than 5 u round down. then u replace the 43 with 0's
Answer:
I am confused as to what you're asking.
Answer:
ins the same
Explanation:
neutrons have no charge
Answer:
<h3>The answer is 320.75 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>320.75 mL</h3>
Hope this helps you
Plant: leaf tissue Human: lungs