Question

In a laboratory experiment, a student titrates 10.0 mL of acetic acid, CH3COOH, solution with 0.5 M NaOH. The endpoint was reached when 46.00 mL of 0.50 M NaOH were delivery. What was the concentration of acetic acid in the solution

Answer 1

Answer:

Concentration of acetic acid=2.3mol/L

Explanation:

Write the balanced chemical reacion:

$CH_3COOH +NaOH = Na(CH_3COO) +H_2O$

N1 and N2 normality of CH3COOH and NaOH respectively;

V1 and V2 volume of CH3COOH and NaOH respectively and they are taken upto end point of titration;

we have given molarity but we need normality;

$Normality =molarity \times n-factor$

but in case of NaOH and CH3COOH n-factor is 1 for each.

hence

normality=molarity;

$N_1V_1=N_2V_2$

$N_1 \times 10 =0.5\times 46$

$N_1=2.3=molarity=concentration;$

Concentration of acetic acid=2.3mol/L