2) b. During the titration of HClO4 solution with 0.1032 M NaOH, quantitative an

Question

2) b.

During the titration of HClO4 solution with 0.1032 M NaOH, quantitative analysis student Alchol became distracted and overshot the end point. A fellow student , I.   M. Smart, suggested that he simply record the present volume of NaOH added and titrate the excess with standard acid solution. If the orginal sample volume was 25 mL, the volume of NaOH added was 28.06 mL, and it took 3.47 mL of 0.1094 M HCl to back-titrate the NaOH, calculate the molar concentration of the orginal HClO4 solution.  

Explanation / Answer

M = 0.1032 NaOH

Voriginal = 25 mL

Vtotal = 28.06 mL

V = 3.47 mL of HClO4 extra

M = 0.1094 M of HClO4

total mol of OH used:

mol = MV =0.1032 *28.06 = 2.895792 mmol of OH-

mmol of HClO4 used in extra titration = MV = 0.1094*3.47 = 0.379618 mmol

then

mmol of OH- acutally used in first titration = 2.895792 -0.379618 = 2.516174 mmol of OH-

so

mmol of HClO4 initially = 2.516174

[HClO4 ] initial = mmol/mL = 2.516174/25 = 0.10064696 M

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