1) A 0.23 g sample of an unknown monoprotic acid is titrated with 0.11 N NaOH. 2

Question

1) A 0.23 g sample of an unknown monoprotic acid is titrated with 0.11 N NaOH. 23.89 mL of NaOH is required to reach the equivalence point. What is the molar mass of this acid?

2) You perform a phenolphthalein titration on a 0.26 g sample of an unknown organic acid, which requires 33.78 mL of 0.1008 N NaOH to reach the endpoint.

a) What is the equivalent weight (estimate) for this acid?

b) Assume the acid is monoprotic and calculate its molar mass. Assume the acid is diprotic and calculate its molar mass. Now that you have two potential molar mass values, using the list of acids in the lab, do you think this acid is monoprotic or diprotic? Explain your rationale.

Explanation / Answer

1) Let the molar mass of the acid = 'M' g/mole

SSince the acid is monoprotic, therefore, molar mass of the acid = equivalent mass of the acid = 'M' g/equivalent

Now, equivalents of NaOH used to reach the point of equivalence = Normality*volume in litres = 0.11*0.02389 = 0.00263

Thus, equivalents of acid present = equivalents of NaOH = 0.00263

Now, equivalents of acid = mass of acid/equivalent ,mass of acid = 0.23/M

or, 0.23/M = 0.00263

or, M = 87.52 g = molar mass of the acid

2) equivalents of NaOH used to reach the point of equivalence = Normality*volume in litres = 0.1008*0.03378 = 0.00341

Thus, equivalents of acid present = equivalents of NaOH = 0.00341

Now, equivalents of acid = mass of acid/equivalent mass of acid = 0.26/M

or, 0.26/M = 0.00341

or, M = 76.36 g = equivalent mass of the acid

b) As acid is monoprotic, therefore its molar mass = its equivalent mass = 76.36 g/mole

If the acid is diprotic, then equivalent mass = molar mass/2

or, molar mass = 2*equivalent mass = 152.72 g/mole

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