A student was performing a titration of 25.96 mL of 0.4940 mol/L CH_3COOH, aceti

Question

A student was performing a titration of 25.96 mL of 0.4940 mol/L CH_3COOH, acetic acid, with a NaOH solution, sodium hydroxide, of unknown concentration. The student carefully washed the burette and rinsed it with deionized water, but neglected to coat the sides of the burette with the solution before filling it for the titration. The titration required 25.24 mL of the 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 moI/L

Explanation / Answer

At the equivalence

mmoles of acid = mmoles of base

25.96mL x 0.4940 = 25.24 x M

Thus molarity of NaOH = 0.5080 M

the calculated molarity = 0.5080 M

Now since the burette is rinsed with water and is not coated with NaOH solution. some water is tstill there in the burette and diluted the NaOH filled.

Hence the molarity we calculated is less than the actual molarity of NaOH.

Thus the true molarity is more than the calculated 0.5080M

option e.

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