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A compound is known to have a free amino group with a pKa of 8.8, and one other ionizable group with a pKa between 5 and 7. To 100 mL of a 0.2 M solution of this compound at pH 8.2 was added 40 mL of a solution of 0.2 M hydrochloric acid. The pH changed to 6.2. The pKa of the second ionizable group is: A) The pH cannot be determined from this information. B) 5.4. C) 5.6. D) 6.0. E) 6.2.

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Answer:

C) 5.6

Explanation:

From Henderson-Hasselbalch equation:

[tex]pH=pKA+log(\frac{A}{HA})\\ \\8.2=8.8+log(\frac{A}{HA})\\\\log(\frac{A}{HA})=8.2-8.8\\\\log(\frac{A}{HA})=-0.6\\\\\frac{A}{HA}=10^{-0.6}\\\\\frac{A}{HA}=0.251\\[/tex]

A = Conjugate base, HA = Acid

%A = 0.251 / (0.251 + 0.1) = 20%, hence %HA = 80%

100 mL of a 0.2 M = 0.02 mol

base = 20% of 0.02 = 0.004

Acid = 80% of 0.02 = 0.016

0.004 moles of base will get protonated by .004 moles of H+ from HCl, forming 0.004 moles of acid

0.02mol original acid - .004 mol H+ = 0.016 mol of conjugate base

Hence A/HA = 0.016/0.004

Since the new pH is 6.2, therefore:

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