7.36g of a copper salt was dissolved and reacted with iodide to yield solid copp

Question

7.36g of a copper salt was dissolved and reacted with iodide to yield solid copper iodide (CuI) and the triiodide ion I3-(aq). The copper iodide was filtered off and the filtrate containing the I3-(aq) was titrated against a standardised sodium thiosulphate (0.01 mol dm-3). A titre of 26.35 cm3 of thiosulphate was recorded.

(i) Give a balanced equation for the reaction of Cu2+ with I- ...............................(2 marks)

(ii) Give a balanced equation for the reaction I3- with thiosulphate ions (S2O32-). ..................(2 marks)

(iii) Calculate the percentage copper in the original copper salt.................................(4 marks)

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Explanation / Answer

(i) 2Cu2+ (aq) + 5I- (aq)---> 2CuI (s) + I3 –

(ii) 2 S2O32– + I3– ---> S4O62– + 3 I–

(iii) Combining equations (i) and (ii), the two step process can be combined to form this overall equation:

2Cu2+ (aq) + 2 S2O32– +2I- (aq)---> 2CuI (s) + S4O62–

Moles of S2O32– needed for titration = (0.01 * 26.35/1000) mol = 2.635 * 10^-4

Since, the mol ratio of Cu2+ reacting with S2O32– is 1:1

Hence, amount of Cu2+ titrated = 2.635 * 10^-4 mol = 2.635 * 10^-4 * 63.55 g = 0.0167 g

So, 0.0167 g Cu2+ remains in 7.36g of the copper salt

i.e. percentage copper in the original copper salt = (0.0167/7.36) * 100% =0.227%

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