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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

Respuesta :

Answer:

Concentration of acetic acid=2.3mol/L

Explanation:

Write the balanced chemical reacion:

[tex]CH_3COOH +NaOH = Na(CH_3COO) +H_2O[/tex]

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;

[tex]Normality =molarity \times n-factor[/tex]

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

hence

normality=molarity;

[tex]N_1V_1=N_2V_2[/tex]

[tex]N_1 \times 10 =0.5\times 46[/tex]

[tex]N_1=2.3=molarity=concentration;[/tex]

Concentration of acetic acid=2.3mol/L

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