A student is determining the identity of an unknown organic acid. The student pl

Question

A student is determining the identity of an unknown organic acid. The student places .1352 g of the acid in a flask and titrates the sample with 0.075M NaOH. The student uses 11.00 ml of 0.075M NaOH to complete the titration.

A.) How many moles of acid were in the 0.1352 g sample?

B.) what is the equivalent weight of Sample A? (equivalent weight = g solid/mol H+)

**If you could help my by showing me how to set the problem up, that would be great. I know you have to start with the 0.1352g when setting up the conversions but I don't know where to go from there. I also know that the molar mass of NaOH is 39.998, or 40.0.*** Thanks a ton for the help!!!!

Explanation / Answer

(A)

By definition of molarity,

1000 mL 1 M = 1 mole

11 mL 0.075M NaOH = 0.075*11*1/1000 = 0.0008 mole.

So at equivalence point, 0.1352 g of unknown acid = 0.0008 mole

1 Mole of unknown acid = 0.1352/0.0008 = 169 g

(B)

Considering it is a monobasic acid, the equivalent weight = 169 g

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