A student was performing a titration of 25.96 mL of 0.4940 mol/L CH3COOH, acetic

Question

A student was performing a titration of 25.96 mL of 0.4940 mol/L CH3COOH, acetic acid, with a NaOH solution, sodium hydroxide, of unknown concentration. The student carefully washed the burette and rinsed it deionized water, but neglected to coat the sides of the burette with the NaOH solution before filling it for the titration. The titration required 25.24 mL of the NaOH solution. The true concentration of the NaOH would be :

A) exactly 0.2540 mol/L

B) more than the calculated 0.2540 mol/L

C) less than the calculated 0.5080 mol/L

D) exactly 0.5080 mol/L

E) more than the calculated 0.5080 mol/L

Explanation / Answer

The answer is:

E) more than the calculated 0.5080 mol/L


NaOH + CH3COOH => CH3COONa + H2O

Moles of NaOH = moles of CH3COOH = volume x concentration of CH3COOH

= 25.96/1000 x 0.4940 = 0.012824 mol


Calculated concentration of NaOH = moles/apparent volume of NaOH

= 0.012824/0.02524 = 0.5080 mol/L


Since apparent volume of NaOH = true volume of NaOH + excess water

=> true volume of NaOH < apparent volume of NaOH

=> actual concentration = moles/true volume > calculated concentration of 0.5080 mol/L

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