The titration of 10.00 mL of a diprotic acid solution of unknown concentration r

Question

The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution? 10.00 mL of vibega (mass = 10.0g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar. A 0.2602 g sample of an unknown momoprotic acid requires 12.23 mL of 0.1298 M NaOH solution to reach the end point. what is the molecular mass of the acid?

Explanation / Answer

Solution :-

Q4) Given data 10.00 ml diprotic acid

21.37 ml of 0.1432 M NaOH

Molarity of acid = ?

Lets calculate the moles of NaOH

Moles of NaOH = molarity * volume in liter

                           = 0.1432 mol per L * 0.02137 L

                          = 0.003060 mol

Now lets calculate the moles of acid

Moles of Acid = 0.003060 mol NaOH * 1 mol Acid /2 mol NaOH

                         = 0.001530 mol

Now lets find molarity of acid

Molarity of acid = moles / volume of acid

                            = 0.001530 mol / 0.010 L

                            = 0.1530 M

Q5) Given

10 ml vinegar (10.05 g)

14.77 ml of 0.4926 M NaOH

Moles of NaOH = molarity * volume

                           = 0.4926 mol per L * 0.01477 L

                      = 0.007278 mol

Mole ratio of the acetic acid to NaOH is 1 : 1 so the moles of acetic acid are same as moles of NaOH

So the molarity of the acetic acid in vinegar is

Molarity = moles / volume

                = 0.007278 mol / 0.010 L

                = 0.7278 M

Mass of acetic acid = moles * molar mass

                                 =0.007278 mol * 60.05 g per mol

                                 = 0.4370 g

% of acetic acid in vinegar = ( mass of acetic acid / mass of vinegar )*100%

                                            = (0.4370 g/10.05g)*100%

                                            = 4.35 %

Q6) 0.2602 g unknown acid

12.23 ml of 0.1298 M NaOH

Moles of NaOH = molarity * volume in liter

                            = 0.1298 mol per L * 0.01223 L

                            = 0.001587 mol

Moles of acid are same as moles of NaOH since acid is monoprotic

So the molar mass of acid = mass / moles

                                            = 0.2602 g / 0.001587 mol

                                            = 163.9 g per mol

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