Youve been asked by your research advisor to prepare 250.00 mL of a 0.020 M buffer. The target pH...

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Youve been asked by your research advisor to prepare 250.00 mL of a 0.020 M buffer. The target pH is 5.60. The components are

Youve been asked by your research advisor to prepare 250.00 mL of a 0.020 M buffer. The target pH is 5.60. The components are

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Answer 1

Answer #1

Part A : Volume of C₃H₅O₂^- = 83.8 mL

Part B : Volume of HC₃H₅O₂ = 16.2 mL

Explanation

According to Henderson Hasselbalch equation

pH = pKa + log([conjugate base] / [weak acid])

pH = pKa + log([C₃H₅O₂^-] / [HC₃H₅O₂])

5.60 = 4.886 + log([C₃H₅O₂^-] / [HC₃H₅O₂])

log([C₃H₅O₂^-] / [HC₃H₅O₂]) = 5.60 - 4.886

log([C₃H₅O₂^-] / [HC₃H₅O₂]) = 0.714

[C₃H₅O₂^-] / [HC₃H₅O₂] = 10^(0.714)

[C₃H₅O₂^-] / [HC₃H₅O₂] = 5.1754 ...(1)

Also, [C₃H₅O₂^-] + [HC₃H₅O₂] = 0.020 M ...(2)

Solving equation (1) and (2)

[C₃H₅O₂^-] = 0.01676 M

[HC₃H₅O₂] = 0.00324 M

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Answer 2

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