Question #bd074

#sf(8(a). (i))#

Potassium iodide solution could be identified by its reaction with iron(III) sulfate solution.

This is a redox reaction where #sf(I^-)# is oxidised to #sf(I_2)# by the #sf(Fe^(3+))# ions.

Omitting the spectator ions:

#sf(Fe_((aq))^(3+)+I_((aq))^(-)rarrFe_((aq))^(2+)+(1)/2I_(2(aq)))#

Addition of starch solution would give a blue/black precipitate to confirm the presence of iodine since iron(III) and iodine are both brown in colour.

#sf(---------------------)#

Barium chloride solution would give a white precipitate with iron(III) sulfate solution of barium sulfate. Again, omitting the spectators:

#sf(Ba_((aq))^(2+)+SO_(4(aq))^(2-)rarrBaSO_(4(s)))#

Strictly speaking you should add an XS of dilute HCl to dissolve other possible precipitates.

#sf(---------------------)#

If you mix iron(III) sulfate solution with the #sf(K_4Fe(CN)_6)# solution you get a dark blue solid commonly known as "Prussian Blue":

#sf(Fe^(3+)+[Fe^(II)(CN)_6]^(4-)rarr[Fe^(III)[Fe^(II)(CN)_6]]^(-))#

#sf(---------------------)#

#sf((a)(ii))#

Aluminium will dissolve in warm sodium hydroxide solution to give sodium aluminate and hydrogen:

#sf(2Al_((s))+2NaOH_((aq))+2H_2O_((l))rarr2NaAlO_(2(aq))+3H_(2(g)))#

This reflects the amphoteric nature of aluminium.

The same type of reaction occurs with zinc which is also amphoteric:

#sf(Zn(s)+H_2O(l)+2NaOH(aq)rarrNa_2Zn(OH)_2(aq)+H_2(g))#

If ammonium ions are warmed with hydroxide ions, ammonia gas is evolved which will turn red litmus blue:

#sf(NH_(4(aq))^++OH_((aq))^(-)rarrNH_(3(g))+H_2O_((l)))#

An aqueous solution of chlorine will react with sodium hydroxide solution, removing any green colouration due to elemental chlorine:

#sf(Cl_(2(aq))+2NaOH_((aq))rarrNaCl_((aq))+NaOCl_((aq))+H_2O_((l)))#

This is an example of "disproportionation". Chlorine (0) is simultaneously oxidised to #sf(OCl^-)# (+1) and to #sf(Cl^-)# (-1).

#sf((b)(i)#

#sf(A)# is ammonium chromate(VI) and decomposes on heating to give green chromium(III) oxide:

It is quite spectacular and is known as "The Volcano Experiment".

#sf((NH_4)_2Cr_2O_(7(s))rarrCr_2O_(3(s))+N_(2(g))+4H_2O_((g))#

So #sf(M)# is chromium.

#sf(B)# is chromium(III) oxide #sf(Cr_2O_3)#.

#sf(C)# is nitrogen gas #sf(N_2)#.

#sf(N_2)# will burn in magnesium to give magnesium nitride:

#sf(3Mg_((s))+N_(2(g))rarrMg_3N_(2(s))#

So #sf(D)# is magnesium nitride.

This reacts with water:

#sf(Mg_3N_(2(s))+6H_2O_((l))rarrMg(OH)_(2(s))+2NH_3(g))#

So #sf(E)# is ammonia, which turns red litmus paper blue.

When aqueous #sf(A)# is warmed with carbonate ions an acid base reaction occurs giving #sf(E)# which we know is ammonia:

#sf(NH_(4(aq))^++CO_(3(aq))^(2-)rarrNH_(3(g))+HCO_(3(aq))^-)#

Green chromium(III) oxide is also amphoteric and will dissolve in aqueous alkali to give #sf([Cr(OH)_6]^(3-))# which is soluble.

Under these conditions hydrogen peroxide is a powerful oxidising agent and is able to oxidise green #sf(Cr(III))# to give yellow #sf(Cr(VI))#:

#sf(2[Cr(OH)_6]^(3-)+3H_2O_2rarr2CrO_4^(2-)+2OH^(-)+8H_2O)#