# F324 Titration question (but just formulating equations)

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Thread starter 7 years ago
#1
Please can I have help on this question? I'm stressing out about it so much! (Q8 june 2011) http://www.ocr.org.uk/images/65869-q...d-elements.pdf

Step 1: 2.80 g of brass is reacted with an excess of concentrated nitric acid, HNO3.
The half-equations taking place are shown below.
Cu(s) --> Cu2+(aq) + 2e–
2HNO3(l) + e– -->NO3–(aq) + NO2(g) + H2O(l)

Step 2: Excess aqueous sodium carbonate is added to neutralise any acid. The mixture effervesces and a precipitate forms.

Step 3: The precipitate is reacted with ethanoic acid to form a solution which is made up to 250 cm3 with water.

Step 4: A 25.0 cm3 sample of the solution is pipetted into a conical flask and an excess of aqueous potassium iodide is added. A precipitate of copper(I) iodide and a solution of iodine, I2(aq), forms.

Step 5: The resulting mixture is titrated with 0.100 mol dm–3 sodium thiosulfate to estimate the iodine present:

I2(aq) + 2S2O3^2–(aq) --> 2I–(aq) + S4O6^2–(aq)

Step 6 Steps 4 and 5 are repeated to obtain an average titre of 29.8 cm3.

1) For steps 1, 2 and 4 write ionic equations including state symbols for the reactions taking place.---------------------------------------
Step 1 was easy, but with step 2 the answer turns out to be:
Cu2+ + CO3- -> CuCO3
2H+ + CO3^2- -> H2O + CO2
Overall: Cu2+ + 2H+ + 2CO3^2- -> CuCO3 + H2O + CO2
I understand the first part enough as Cu2+ is the product and we're adding the sodium carbonate afterwards (although would never have thought to add it seperately to each half equation provided). However is the reason that we're adding the CO3^2- to the H+ (from the HNO3) which is on the other side of the equation unlike the Cu2+ because the acid is in excess? I know that the question actually says it's neutralising the acid, but how come it's on the other side of the equation? Is it being in excess means we can assume there's also HNO3 on the right hand side of the equation too? (i.e. a product)?

Also, please can someone talk me through how to work out step 4? The answer is supposed to be 2Cu2+ + 4I- -> 2CuI + I2.
I also got confused because I was trying to work out the formula would be of the precipitate and then what that would be when it's reacted with the ethanoic acid, but I guess we can ignore that because it's ionic so it'll still be Cu2+?

Thank you very much and sorry I know there's a lot to read
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7 years ago
#2
Don't worry, an A* student of mine started crying in this exam because of this question.

Step 1 essentially makes the mixture 'safe': conc. nitric is nasty stuff. Two reactions happen: the acid is neutralised then Cu2+ forms a ppt once the acid has gone (else it would react, like metal carbonates do in acid).

Once all the dangerous acid has gone, the CuCO3 is safely (and prettily) dissolved in ethanoic acid (which acts as a ligand) ready to react with I-.

Step 4 requires you to work out two half and one overall equation

I- -> I2 AND Cu2+ -> Cu+.

The whole point of the question is to find out if you can follow the stoichiometry at each step.
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