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. A) 6.8 B) 0.15 C) 0.21 D) 0.42 E) 130

Question

Quizplus · Accepted Answer

The neutralization reaction is CH3CO2H + NaOH → CH3CO2Na + H2O. Using the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of acetic acid, and M2 and V2 are the molarity and volume of NaOH, we find M1 = (0.250 M * 29.6 mL) / 17.5 mL = 0.42 M.

A 175 ML Sample of an Acetic Acid (CH3CO2H) Solution Required

A 175 ML Sample of an Acetic Acid (CH₃CO₂H) Solution Required