I am having so much mental agony on this so if someone has to correct my maths, please could you show me the steps. So grateful really. Iodometric titration done of copper nitrate with unknown hydrate, is it hexa, deca, or tri or something else?
2.5 g of Copper Nitrate added to water to make up a 100 mL solution
From this mother solution 30 mL was extracted and titrated (dilution factor of 3.33)
80 mL of 0.01 M thiosulphate titrated to the endpoint.
My reasoning:
0.08 L titrant x 0.01 M Thiosulphate = 0.0008 moles of Cu2+ in 30 mL sample
Dilution factor is 3.33 therefore this x 0.0008 = 0.00266 moles in the 100 mL original solution.
0.00266 x 187.55 g/mol (anhydrous copper nitrate) = 0.5 g of Cu(NO3)2 in that 100 mL
Therefore logic dictates that 2.5 g - 0.5 g = 2.0 g of this nitrate sample is water
This means that 2.5/0.5 = 5 which means that my sample is Cu(NO3)2.5H2O
Thankyou