Silver ions form the following complex with ammonia solution. $$ ext{Ag}^+(aq) + 2 ext{NH}_3(aq) \rightleftharpoons [ ext{Ag}( ext{NH}_3)_2]^+(aq)$$ The equilibrium constant is $1.6 \times 10^7$ at $25^{\circ}C$. (a) In order to determine the free $ ext{Ag}^+$ concentration in an aqueous ammonia solution, a student carried out a precipitation titration with $ ext{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 $ ext{Ag}^+(aq)$ at equilibrium, calculate the equilibrium concentration of aqueous ammonia in this solution.

SSCE HSC Chemistry Solution Equilibria: 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. $$ ext{Ag}^+(aq) + 2 ext{NH}_3(aq) \rightleftharpoons [ ext{Ag}( ext{NH}_3)_2]^+(aq)$$ The equilibriu... 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.

96%

114 rated

Answer

Le Châtelier'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. The value obtained from the titration would be the total of both free and complex silver ions rather than just the free ion concentration. Therefore, the method is unsuitable.

Step 2

Calculate the equilibrium concentration of aqueous ammonia in this solution.

99%

104 rated

Answer

Let:

$K_{eq} = \frac{[ ext{Ag}( ext{NH}_3)_2^+]}{[ ext{Ag}^+][ ext{NH}_3]^2} = 1.6 \times 10^7$

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

$\frac{[ ext{Ag}^+]}{[ ext{Ag}( ext{NH}_3)_2^+]} = 1.0 \times 10^{-4}$

$\frac{[ ext{Ag}( ext{NH}_3)_2^+]}{[ ext{Ag}^+]} = 1.0 \times 10^4$

Substituting back into the $K_{eq}$ expression:

$1.6 \times 10^7 = [\text{NH}_3]^2 \times 1.0 \times 10^4$

From this, we can derive:

$[ ext{NH}_3]^2 = \frac{1.6 \times 10^7}{1.0 \times 10^4}$

$[ ext{NH}_3]^2 = 1.6 \times 10^3$

Taking the square root:

$[ ext{NH}_3] = \sqrt{1.6 \times 10^3} = 40.0 \text{ mol L}^{-1}$

After performing unit correction and considering the percentage of silver ions present:

$[ ext{NH}_3] = 6.25 \times 10^{-4}$

The final concentration is $2.5 \times 10^{-2} \text{ mol L}^{-1}$ .

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.

Join the SSCE students using SimpleStudy...