Silver ions form the following complex with ammonia solution. $$Ag^+(aq) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq)$$ The equilibrium constant is $1.6 \times 10^7$ at $25^{\circ}C$. (a) In order to determine the free $Ag^+$ concentration in an aqueous ammonia solution, a student carried out a precipitation titration with NaI(aq) as the titrant. Evaluate the suitability of this method. (b) If 0.010 ext{%} of the total silver ions in solution are present as $Ag^+(aq)$ at equilibrium, calculate the equilibrium concentration of aqueous ammonia in this solution.

SSCE HSC Chemistry Concentration and Molarity: Silver ions form the following c

Silver ions form the following complex with ammonia solution - HSC - SSCE Chemistry - Question 31 - 2022 - Paper 1

Question 31

Silver ions form the following complex with ammonia solution. $$Ag^+(aq) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq)$$ The equilibrium constant is $1.6 \time... show full transcript

Worked Solution & Example Answer:Silver ions form the following complex with ammonia solution - HSC - SSCE Chemistry - Question 31 - 2022 - Paper 1

Step 1

Evaluate the suitability of this method.

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Answer

Le Chatelier's Principle predicts that as the silver ions precipitate, the complex will decompose to release more silver ions. This disturbs the equilibrium to the left.

This process will continue until all of the complex is broken up into its ions, so the value obtained from the titration would be the total of both free and complex silver ions rather than just the free. Therefore the method is unsuitable.

Step 2

Calculate the equilibrium concentration of aqueous ammonia in this solution.

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Answer

The equilibrium constant is given by:

$K_{eq} = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} = 1.6 \times 10^7$

Since the concentration of free silver is very low, assume the ratio of free to complex silver is approximately equal to 0.010\text{%} (or $1.0 \times 10^{-4}$ ):

$\frac{[Ag^+]}{[Ag(NH_3)_2^+]} = 1.0 \times 10^{-4}$

From the equilibrium expression, we have: $[Ag(NH_3)_2^+] = \frac{1.6 \times 10^7 \cdot [NH_3]^2}{[Ag^+]}$

We substitute: $1.6 \times 10^7 \cdot [NH_3]^2 = 1.0 \times 10^4$

Solving for $[NH_3]^2$ gives: $[NH_3]^2 = 6.25 \times 10^{-4}$

Thus, $[NH_3] = 2.5 \times 10^{-2} mol \cdot L^{-1}$

Silver ions form the following complex with ammonia solution - HSC - SSCE Chemistry - Question 31 - 2022 - Paper 1

Worked Solution & Example Answer:Silver ions form the following complex with ammonia solution - HSC - SSCE Chemistry - Question 31 - 2022 - Paper 1

Evaluate the suitability of this method.

Calculate the equilibrium concentration of aqueous ammonia in this solution.

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