Titration Chemical Calculation

Question

Solution A contains 6.30g/dm3 of a dibasic acid, H2C2O4.xH2O (where x is a real number). Solution B contains 0.100 mol/dm3 of sodium hydroxide.

Solution A was titrated with solution B and it was found that 20.0 cm3 of A required 19.5 cm3 of B for complete reaction.

If 1 mole of the acid A reacts with 2 moles of B, find the value of x.

Answer 1

Moles NaOH = 0.100 mol/dm^3 x 0.0195 dm^3 =0.00195

Moles acid = 0.00195/2 =0.000975

mass acid = 0.000975 mol x 90.03 g/mol =0.0878 g in 20 cm^3 => 0.020 dm^3

0.0878 g : 0.020 dm^3 = x : 1 dm^3

x = 4.39 g of acid in 1 dm^3

mass water = 6.30 - 4.39 = 1.91 g

moles water = 1.91 g / 18.02 g/mol = 0.106

Moles acid = 4.39 g/ 90.03 g/mol =0.0488

we get

H2C2O4 ( 0.0488) H2O (0.106)

we divide by the smallest number :

0.0448/0.0488 = 1 => C2H4O2

0.106 / 0.0488 = 2 => H2O

x = 2

C2H4O2 . 2H2O

it is oxalic acid dihydrate