Question bd074

Jul 7, 2016

See below:

Explanation:

$\textsf{8 \left(a\right) . \left(i\right)}$

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

This is a redox reaction where $\textsf{{I}^{-}}$ is oxidised to $\textsf{{I}_{2}}$ by the $\textsf{F {e}^{3 +}}$ ions.

Omitting the spectator ions:

$\textsf{F {e}_{\left(a q\right)}^{3 +} + {I}_{\left(a q\right)}^{-} \rightarrow F {e}_{\left(a q\right)}^{2 +} + \frac{1}{2} {I}_{2 \left(a q\right)}}$

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.

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

$\textsf{B {a}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -} \rightarrow B a S {O}_{4 \left(s\right)}}$

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

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If you mix iron(III) sulfate solution with the $\textsf{{K}_{4} F e {\left(C N\right)}_{6}}$ solution you get a dark blue solid commonly known as "Prussian Blue":

$\textsf{F {e}^{3 +} + {\left[F {e}^{I I} {\left(C N\right)}_{6}\right]}^{4 -} \rightarrow {\left[F {e}^{I I I} \left[F {e}^{I I} {\left(C N\right)}_{6}\right]\right]}^{-}}$

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$\textsf{\left(a\right) \left(i i\right)}$

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

$\textsf{2 A {l}_{\left(s\right)} + 2 N a O {H}_{\left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)} \rightarrow 2 N a A l {O}_{2 \left(a q\right)} + 3 {H}_{2 \left(g\right)}}$

This reflects the amphoteric nature of aluminium.

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

$\textsf{Z n \left(s\right) + {H}_{2} O \left(l\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} Z n {\left(O H\right)}_{2} \left(a q\right) + {H}_{2} \left(g\right)}$

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

$\textsf{N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} \rightarrow N {H}_{3 \left(g\right)} + {H}_{2} {O}_{\left(l\right)}}$

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

$\textsf{C {l}_{2 \left(a q\right)} + 2 N a O {H}_{\left(a q\right)} \rightarrow N a C {l}_{\left(a q\right)} + N a O C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}}$

This is an example of "disproportionation". Chlorine (0) is simultaneously oxidised to $\textsf{O C {l}^{-}}$ (+1) and to $\textsf{C {l}^{-}}$ (-1).

sf((b)(i)

$\textsf{A}$ 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 $\textsf{M}$ is chromium.

$\textsf{B}$ is chromium(III) oxide $\textsf{C {r}_{2} {O}_{3}}$.

$\textsf{C}$ is nitrogen gas $\textsf{{N}_{2}}$.

$\textsf{{N}_{2}}$ will burn in magnesium to give magnesium nitride:

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

So $\textsf{D}$ is magnesium nitride.

This reacts with water:

$\textsf{M {g}_{3} {N}_{2 \left(s\right)} + 6 {H}_{2} {O}_{\left(l\right)} \rightarrow M g {\left(O H\right)}_{2 \left(s\right)} + 2 N {H}_{3} \left(g\right)}$

So $\textsf{E}$ is ammonia, which turns red litmus paper blue.

When aqueous $\textsf{A}$ is warmed with carbonate ions an acid base reaction occurs giving $\textsf{E}$ which we know is ammonia:

$\textsf{N {H}_{4 \left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -} \rightarrow N {H}_{3 \left(g\right)} + H C {O}_{3 \left(a q\right)}^{-}}$

Green chromium(III) oxide is also amphoteric and will dissolve in aqueous alkali to give $\textsf{{\left[C r {\left(O H\right)}_{6}\right]}^{3 -}}$ which is soluble.

Under these conditions hydrogen peroxide is a powerful oxidising agent and is able to oxidise green $\textsf{C r \left(I I I\right)}$ to give yellow $\textsf{C r \left(V I\right)}$:

$\textsf{2 {\left[C r {\left(O H\right)}_{6}\right]}^{3 -} + 3 {H}_{2} {O}_{2} \rightarrow 2 C r {O}_{4}^{2 -} + 2 O {H}^{-} + 8 {H}_{2} O}$