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Alex787 [66]
2 years ago
11

Why can tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addit

ion of sodium hydroxide
Chemistry
1 answer:
devlian [24]2 years ago
7 0

Answer:

Tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide because they become soluble whereas copper (II) sulfide and bismuth (III) sulfide remain insoluble.

Explanation:

Sodium hydroxide is a  basic solution which is used as a precipitating agent for metallic ions in the laboratory.

<em>When a solution containing a mixture of the sulfides of the Group II cations,  antimony (III), copper (II), tin  (IV), and bismuth (III), is made basic by the addition of a base such as sodium hydroxide or ammonium hydroxide, the  sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become  soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in  water, while the sulfides of copper (II) and bismuth (III) do not.</em> The result is the dissolution of the antimony (III) sulfide  and tin (IV) sulfide, separating them from the copper (II) sulfide and the bismuth (III)  sulfide.

Sb₂S₃(s) + 3S²⁻(aq) ----> 2SbS₃³⁻(aq)

SnS₂(s) + S²⁻(aq) ----> SnS₃³⁻(aq)

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Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

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