You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.050 M HOAc and 0.050 M NaOAc. How do the pHs of the buffered solutions compare? (Note: Ac - = acetate ion, CH3COO-). A. The pH of buffered solution 1 is greater than that of buffered solution 2. B. None of the answers are correct C. The pH of buffered solution 1 is equal to that of buffered solution 2. D. The pH of buffered solution 2 is greater than that of buffered solution 1. E. Cannot be determined without the Ka values.

Question

Aarav Wagner

Chemistry

6 July, 02:31

You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.050 M HOAc and 0.050 M NaOAc. How do the pHs of the buffered solutions compare? (Note: Ac - = acetate ion, CH3COO-). A. The pH of buffered solution 1 is greater than that of buffered solution 2. B. None of the answers are correct C. The pH of buffered solution 1 is equal to that of buffered solution 2. D. The pH of buffered solution 2 is greater than that of buffered solution 1. E. Cannot be determined without the Ka values.

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Answers (1)

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Makaila Jefferson · Answer 1

C

Explanation:

The Henderson-Hasselbalch equation relates pH to the concentrations of an weak acid-base conjugate pair as follows:

pH = pKa + log ([A⁻]/[HA])

For solution 1, the pH may be expressed as follows:

pH = pKa + log (5.0M/5.0M) = pKa

For solution 2, pH may be expressed as follows:

pH = pKa + log (0.050M/0.050M) = pKa

Thus, the pH values are equal to the pKa in both cases and are the same.