A 17.5 ml sample of an acetic acid (ch3co2h) solution required 29.6 ml of 0.250 m naoh for neutralization. the concentration of acetic acid was m.

0.423M

Explanation:

The reaction of acetic acid with NaOH is:

CH₃CO₂H + NaOH → CH₃CO₂Na + H₂O

That means the moles of NaOH added = Moles acetic acid in the solution

To solve this problem we need to find the moles of NaOH added to find the moles of acetic acid and its concentration using its volume as follows:

Moles NaOH:

29.6mL = 0.0296L * (0.250mol / L)= 0.0074 moles NaOH = Moles acetic acid

Molarity acetic acid:

0.0074 moles / 0.0175L =

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