The titration of 25.00 mL of a sulfuric acid solution of unknown concentration r

Question

The titration of 25.00 mL of a sulfuric acid solution of unknown concentration requires 31.22 mL of a 0.1234 M NaOH solution What is the concentration of the sulfuric acid solution? 10.00 mL of vinegar (mass = 10.05 g) requires 16.28 mL of 0.5120 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar. A 0.1936 g sample of an unknown monoprotic acid requires 15.56 mL of 0.1020 M NaOH solution to reach the end point. what is the molecular mass of the acid?

Explanation / Answer

1) H2SO4(aq) + 2NaOH(aq) ------> Na2SO4(aq) + 2H2O

As per the balanced reaction H2SO4 & NaOH reacts in the molar ratio of 1:2

Thus, moles of H2SO4 reacting = 2*moles of NaOH = 2*0.1234*0.03122 = 7.705*10-3

Thus, H2SO4 concentration = moles of H2SO4/Volume of litres = 0.308 M

2) CH3COOH(aq) + NaOH(aq) ---------> CH3COONa(aq) + H2O(l)

As per the balanced reaction CH3COOH & NaOH reacts in the molar ratio of 1:1

THus, molarity of CH3COOH = molarity of NaOH*volume of NaOH solution in ml/volume of vinegar solution in ml = (0.5120*16.28)/10 = 0.834 M

moles of acetic acid = molarity*volume of solution in litres = 0.834*0.01 = 0.00834

molar mass of acetic acid = 60 g/mole

Thus, mass of acetic acid = moles*molar mass = 0.5 g

mass % = (mass of acetic acid/mass of vinegar)*100 = 5 %

3) Since the acid is monoprotic, therefore moles of the acid present = moles of NaOH = molarity of NaOH*volume of NaOH solution in litres = 0.102*0.01556 = 1.587*10-3

Now, moles of acid = mass of acid/molar mass of the acid

or, molar mass of the acid = mass/moles of the acid = 121.98 g/mole

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