6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added…
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Question “6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added…”
6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added, the pH of the resulting solution was 6.70. After 50.0 mL of NaOH was added, the pH of the solution was 8.00. What are the values of Ka1 and Ka2?
Answer
Below are the steps and complete acid-base reactions of H 2A with NaOH.
H 2A aq + NaOH aq ——-> NaHA aq + H 2O l
NaHA (aq), + NaOH (aq), Na A (aq), + H O (l)
The overall acid-base equation can be found here
H 2A (+ 2 NaOH) ——-> Na 2A(aq), + 2 H 2O
Note the ionic equations used in the ionization reactions.
H2A (aq) + OH- (aq) ——–> HA- (aq) + H2O (l) pKa1
HA- (aq) + OH- (aq) ——–> A2- (aq) + H2O (l) pKa2
Millimols H 2A = (50.0mL)*(0.10M) = 5.0mmol.
Millimols NaOH = 25.0mL = (25.0mL)*(0.10M) = 2.5mmol
Millimols NaOH = 50.0 mL = (50.0mL)*(0.10M) = 5.0mmol.
Add 25.0 mL NaOH to 50.0mL H 2A and half of H 2A to HA –, and the remaining H 2A is left in the solution.
The Henderson-Hasslebach equation gives the pH of the solution.
pH = pKa1 + log [HA–]/[H2A]
= pK a1 [HA –] = H 2A] because the original H 2A is only half neutralized.
Because pH = 6.70,
pKa1 = 6.70
=====> K antilog (-6.70).
=====> Ka1 = 1.995*10-7 2.00*10-7 (ans).
Add 50.0mL 0.10M NaOH to 50.0mL 0.10M H 2A and H 2A to create HA –. HA – is an amphoteric solution. The pH of the solution can be found as follows:
pH = 1/2 (pK A1 +pK A2).
======> 8.00 = 1/2 *(6.70 + pKa2)
======> 16.00 = 6.70 + pKa2
======> pKa2 = 9.30
======> K antilog (-9.30).
======> Ka2 = 5.012*10-10 5.01*10-10 (ans).
Conclusion
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