A buffer solution with a pH of 5.28 is prepared with the same volumes of 0.60 M HC2H3O2 and M NaC2H3O2. The Ka of HC2H3O2 is 1.8 ⋅ 10-5.

Molar concentration of [NaC₂H₃O₂]  is 2.08M

Explanation:

It is possible to find pH of the HC₂H₃O₂ / C₂H₃O₂⁻ buffer using H-H equation, thus:

pH = pKa + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]

Where:

pH = 5.28

pKa is -log Ka: 4.74

[HC₂H₃O₂] = 0.60M -It is possible to use the 0.60M concentration because you will add the same volume of both [HC₂H₃O₂] and [NaC₂H₃O₂] solutions.-

[C₂H₃O₂⁻] = ?

Replacing:

5.28 = 4.74 + log₁₀ [C₂H₃O₂⁻] / [0.60]

0.54 = log₁₀ [C₂H₃O₂⁻] / [0.60]

3.47 =  [C₂H₃O₂⁻] / [0.60]

2.08M = [NaC₂H₃O₂]

Molar concentration of [NaC₂H₃O₂]  is 2.08M

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

c

explanation: