Given that the zinc ions in the ‘Brass Contaminated Sand (BCS)’ solution do not

Question

Given that the zinc ions in the ‘Brass Contaminated Sand (BCS)’ solution do not react with thiosulphate and can be considered as spectator ions. Assume that the reaction between copper ions and thiosulphate ions is a 1 to 1 reaction.
Note: The assumption is a correct statement but is an over-simplification of the overall reaction. The assumption is given to allow you to undertake the calculation without knowledge of the reaction.

Calculate the Molarity of copper in the ‘Brass Contaminated Sand (BCS)’ solution and quote your answer to 4 decimal places.( Molarity of Na2S2O3.5H2O =0.07999 M and the Volume 25ml ,20 ml of Brass contaminated sand used.
And Calculate the % w/w of copper in the brass component of the Brass Contaminated Sand (BCS) solid and quote your answer to 2 decimal placesA sample of Brass Contaminated Sand (BCS) from the fly tip site is found to contain 3.87 % w/w brass and has been quantitatively prepared for you by weighing accurately 1280.8199 gram of Brass Contaminated Sand (BCS )

Explanation / Answer

Spectator ions are those which are simply present in the solution but donot react with any other species.

We are given in the question that Copper and thiosulphate is a 1 to 1 reaction which means 1 mole of copper reacts with 1 mole of thiosulphate ions(S2O32-).

Given, Molarity of Na2S2O3.5H2O = 0.07999M in 25ml solution

Volume of Brass Contaminated Sand (BCS) added to react with 25ml solution of thiosulphate ions = 20ml.

Molarity = no of moles of solute/Volume of solution (in litre)

0.07999 = (no of moles of Na2S2O3.5H2O) / (25 X 10-3)        [10-3 is multiplied to convert ml into Litre]

no of moles of Na2S2O3.5H2O = 0.07999 X 25 X 10-3 = 1.99975 X 10-3

1 Mole Na2S2O3.5H2O = 1 mole S2O32-

Hence, no of moles of thiosulphate ions (S2O32-) = 1.99975 X 10-3

1.99975 X 10-3 moles of thiosulphate ions react with 1.99975 X 10-3 moles of copper.

Molarity of copper in Brass contaminated sand = no of moles of copper/ volume of BCS(solution) in Litre

                                                                           = (1.99975 X 10-3 ) / (20 X 10-3)

                                                                           = 9.99875 X 10-2 M

Rounding off till 4 decimal points = 9.9988 X 10-2 M

%w/w is mass of solute in gram/ mass of solution(or overall mass) in gram. For example - there is 2gram copper in 10 gram brass then w/w of copper is 2/10 = 0.2 and %w/w is 0.2 X 100 = 20%

Given, we have 3.87% w/w brass in 1280.8199gram BCS.

Hence, (mass of brass / 1280.8199) X 100 = 3.87

mass of brass = (3.87 X 1280.8199) / 100

                       = 49.56773013

%w/w of copper in brass component of BCS = (mass of copper in gram / 49.56773013) X 100   [after calculating this round off the answer till two decimal places]

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