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)0.296 B)0.423 C)0.591 D)0.105 E)129.5

Question

Quizplus · Accepted Answer

To find the concentration of acetic acid, we use the neutralization reaction equation and the molarity (M) equation. The neutralization reaction between acetic acid (CH3COOH) and NaOH is 1:1. Thus, moles of NaOH = moles of CH3COOH. Moles of NaOH = 0.250 M * 0.0296 L = 0.0074 moles. To find the concentration of acetic acid: moles of CH3COOH = 0.0074 moles, volume of acetic acid solution = 0.0175 L. Concentration of acetic acid = moles/volume = 0.0074 moles / 0.0175 L = 0.423 M.

A 175 ML Sample of an Acetic Acid (CH3CO2H)solution Required 29

A 175 ML Sample of an Acetic Acid (CH₃CO₂H)solution Required 29