## Acid/Base fun (not really)

**Anonymous**

The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126 grams of HX was dissolved in distilled water and the volume brought to exactly 250.00 mL. Several 50.00 mL portions of this sol'n were titrated against NaOH sol'n, requiring an average of 38.21 mL of NaOH.

The NaOH sol'n was standardized against oxalic acid dihydrate (H2C204x2H20 mole.wt: 126.066) The volume of NaOH sol'n required to neutralize 1.2596 g of oxalic acid was 41.24 mL.

a)Calculate the molarity of the NaOH sol'n

b)Calculate the number of moles of HX in a 50.00 mL portion used for titration

c)Calculate the molecular weight of HX

d)Discuss the effect of the calculate molecular weight of HX if the sample of oxalic acid dihydrate contained a nonacidic impurity.

That's all. I know, it's pretty ugly, but any help would be appreciated =)

Anonymous| 27 April 2005 - 6:45pmHey, i think i have solved this titration question.

a) Use the info given in the standardization question:

(.2596 g)/(126.066 g/mol) * 1/(0.04124 L) = 0.2407 M

b) Use the molarity of NaOH found fo find the moles of NaOH wich is equivalent to the moles of HX in 50 mL, and the fact that 38.21 mL of NaOH was used.

(0.2407 M) * (38.21 mL / 1000) = 9.197e-3 mol of HX in 50 mL of solution

c) you know the moles, and the amount used in grams, you just need to apply the dilution factor.

(15.26 g) / [9.197e-3 mol * (250 mL / 50 mL)] = 331.73 g/mol

d) if ther was an impurity the calculated molarity of NaOH would be inaccurate, therefore the molecular weight of HX would be inaccurate.

if somone could double check my work, that would be great.. been a while since ive done this