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Chemistry

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A student prepares a lactic acid-sodium lactate buffer solution by mixing 40.mL of 0.50MHC3H5O3(aq) with 200.mL of 1.0M NaC3H5O3(aq) . The pKa of HC3H5O3 is 3.08. What is the pH of the resulting solution?

Answer :

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Answer:

D) 4.08

Explanation:

HC3H5O3, [HA]: 0.04L * 0.5M = 0.24L(M2)

M2 = 0.083M

C3H5O3-, [A-]: 1.0L * 0.2L = 0.24L (M2)

M2 = 0.83M

Use the Henderson-Hasselbalch equation:

log(0.83/0.083) = 1

Therefore pH = 3.08 + 1 = 4.08

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