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)0.158 B)0.423 C)134 D)6.88 E)0.214

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction between acetic acid and NaOH:

CH3COOH + NaOH → NaCH3COO + H2O

Next, use the volume and concentration of NaOH to calculate the number of moles of NaOH used:

29.6 mL NaOH x 0.250 mol NaOH/L = 0.0074 mol NaOH

Since the reaction is a 1:1 ratio between acetic acid and NaOH, the number of moles of acetic acid is also 0.0074 mol.

Finally, use the volume of the acetic acid solution to calculate the concentration:

0.0074 mol CH3COOH / 0.0175 L solution = 0.423 M CH3COOH

Therefore, the concentration of the acetic acid solution is 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