Titration Chemistry Problem?

Question

The molecular weight of a monoprotic acid HX was to be de¬termined. A sample of 15.126 grams of HX was dis¬solved in distilled water and the volume brought to ex¬actly 250.00 millilitres in a volumetric flask. Several 50.00 millilitre por¬tions of this solution were titrated against NaOH solution, re¬quiring an average of 38.21 millilitres of NaOH.

The NaOH solution was standardized against oxalic acid di¬hydrate, H2C2O4.2H2O (molecular weight: 126.066 gram mol-1). The volume of NaOH solution required to neutralize 1.2596 grams of oxalic acid dihydrate was 41.24 millilitres.

(a) Calculate the molarity of the NaOH solution.

(b) Calculate the number of moles of HX in a 50.00 millilitre portion used for titration.

(c) Calculate the molecular weight of HX.

Answer 1

(a) Calculate the molarity of the NaOH solution.

first moles of oxalic:

1.2596 grams @ 126.066 gram/mol = 9.99e-3 moles oxalic

then moles of NaOH

9.99e-3 moles oxalic @ 2 moles NaOH / 1 mol Oxalic = 0.01998 mol NaOH

last molarity of NaOH, using moles in 41.24 mls:

0.01998 mol NaOH / 0.04124 litres = 0.4846 Molar NaOH

your first answer:  0.4846 Molar NaOH

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(b) Calculate the number of moles of HX in a 50.00 millilitre portion used for titration.

Hx & NaOH react 1:1, we find the moles of HX in the 50 ml samples by finding the moles of NaOH equal to them:

0.03821 litres NaOH @ 0.4846 mol / litre = 0.01851 moles of NaOH

your answer:  0.01851 moles of Acid

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(c) Calculate the molecular weight of HX.

if  50 ml samples had 0.01851 moles of Acid, then the full 250 mls of sample had 0.1851 moles of acid

Molar mass:

15.126 / 0.1851 moles = 81.70 g/mol

Your last answer: 81.70 g/mol