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Anit [1.1K]
3 years ago
8

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

nolphthalein equilibrium established with water is Hph(aq)(colorless) + H2O (l) H3O+ (aq) + ph-(aq)(pink or red). You compared the color of the solutions in three test tubes that initially contained 3 mL of 0.1 M ammonium hydroxide and a few drops of phenolphthalein indicator. In the first test tube, you added 1 M NH4Cl dropwise. What color change was observed and what did this color change indicate about the shift in the phenolphthalein equilibrium? a. The solution turned a more intense pink or red color indicating that the phenolphthalein equilibrium shifted to the left, producing more of the pink or red colored Hph.
Chemistry
1 answer:
AnnZ [28]3 years ago
7 0

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

\text{NH}_4\text{Cl} is a salt soluble in water. \text{NH}_4\text{Cl} dissociates into ions completely when dissolved.

\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

There's also an equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

The indicator equilibrium will shift to the right to produce more {\text{H}_3\text{O}}^{+} ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding \text{NH}_4\text{Cl} will add to the concentration of {\text{NH}_4}^{+} ions in the solution. Some of the {\text{NH}_4}^{+} ions will combine with \text{OH}^{-} ions to produce \text{NH}_4\text{OH}.

The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

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Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

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Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

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