Question

A 0.5g sample of cement is added to a flask and is placed in a hot water bath with ~20ml of ethylene glycol where it was mixed with a stir bar for 20 minutes. After 20 minute, the liquid was put through a vacuum filtration to collect the liquid. The filtrate was then transferred to another flask.

To the filtrate, 3.00ml of 0.1M HCl was added. 50ml of 0.005M potassium iodate, and 0.5g of solid potassium iodide was also added and dissolved. Starch indicator was added. It was then titrated with standard 0.005M Na2S2O3 (sodium thiosulfate) solution where the solution turned blue. Experimentally, it was determined that 31.12ml of the thiosulfate solution was required to titrate.

Determine the amount of free lime (CaO) present in the sample.

Reaction equations that may be useful...?

CaO (lime) + 2HCl ===> CaCl2 + H2O || (1) Excess HCl used to form iodine which is then detected by the sodium thiosulfate

6HCl + 5KI + KIO3 ===> 3I2 + 6KCl + 3H2O || (2)

I2 + 2Na2S2O3 ===> 2NaI + Na2S4O6 || (3) Thiosulfate ion reduces iodine into iodide ion

Answer 1

3 cad2 +3t2O 5 KI+KI03 = 312 +6KU +3 H20 NaI 3NA2S406 3 me. CaD 3CaD6HC 6HC 31 2 6 Na2203 = 56.0774X Cao G met Nazs2 O3 mstNaz s20s =3(56-0714)= 28-03879 CaO 31.12x 0.005 mdNa20 31.12m O 005 (M) Naz S203 Now AO00 31.12r0 005 Cal 000 O 0044 Cao Pree of Caon the sauble (O-0044K100-0.8g. O5