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Vanyuwa [196]
3 years ago
6

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 0.05

0 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.
Chemistry
1 answer:
andre [41]3 years ago
7 0

Answer:

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.

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<h3>false ..</h3>

since...

  • NaOH is an Arrhenius base because it dissociates in water to give the hydroxide (OH-) and sodium (Na+) ions.

hope this answer is helpful for you

<h3>mark as brainlist please</h3>
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