1) Calculate the number of grams of oxalic acid dihydrate, H 2 C 2 O 4 .2H 2 O,

Question

1) Calculate the number of grams of oxalic acid dihydrate, H2C2O4.2H2O, that are needed to prepare 250.0 mL of 0.15N solution. (When the crystals are dissolved, the water in the hydrate becomes part of the solvent. The crystals you weigh will include hydration water. Hence, the two moles of water must be included in the molar mass of the acid when you make your calculations.)

2) How many milliliters of 6.0 NaOH must be dilited to 500. milliliters to prepare a 0.15N solution?

H2C2O4 + 2NaOH yields Na2C2O4 + 2H2O   (I don't know if this is needed but it was given in the lab)

Explanation / Answer

1) The equivalent weight of oxalic acid dihydrate is equal to half the molecular weight since oxalic acid is dibasic.

Molecular weight of oxalic acid dihydrate is 126, so its equivalent weight is half that or 63.

If you want 250 ml. (0.25 liter) of 0.15 N solution you need to dissolve 0.25 times 0.15 N of its equivalent weight
or 0.0375.

So The weight of oxalic acid required to prepare 250 ml of 0.15N solution is 0.0375 x 63 = 2.36 g.

2) We know N1 V1 = N2 V2 is a formula used to calculate volume required of a particular concentration to prepare know volume of another concentration.

6.0 x X = 0.15 x 500

So X = 12.5 mL

so 12.5 mL of 6.0 N NaOH is required to make 500 mL of 0.15N solution of NaOH

H2C2O4 + 2NaOH yields Na2C2O4 + 2H2O equation is needed to show that oxalic acid is dibasic and hence equivalent weight is 1/2 the molecular weight.

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