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A mixture of sodium hydroxide, naoh(s) and benzoic acid, hc7h5o2(s), consists of a total of 0. 50 moles. This mixture is dissolved in 1. 0 l of water and the ph of the resulting solution is determined to be 12. 52. Find the mole fraction of benzoic acid in the initial mixture

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ogorwyne

With a resulting solution of 12.52 and 1 m.l of water, the mole fraction of Benzoic acid that is in this mixture is 0.4669

How to solve for the mole fraction of Benzoic acid

pH = 12.52

pOH = 14 - 12.52

= 1.48

OH⁻ = [tex]10^-^1^.^4^8[/tex]

= 0.0331

The concentration of excess ions

The total number of excess OH⁻ that is required to make salt

= 0.5 - 0.0331

= 0.4669

H+ = 0.4669/2

= 0.2334

The mole fraction of Benzoic acid that would be here would be gotten by:

0.2334/0.5 = 0.4669

We conclude that the mole of  Benzoic acid would be 0.4669 moles.

Read more on Benzoic acid here: https://brainly.com/question/3460480

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