Commercial silver-plating operations frequently use a solution containing the complex ag+ ion. because the formation constant (kf) is quite large, this procedure ensures that the free ag+ concentration in solution is low for uniform electrodeposition. in one process, a chemist added 9.0 l of 5.0 m nacn to 90.0 l of 0.21 m agno3.

calculate the concentration of free ag+ ions at equilibrium.

see your textbook for kf values.

[Ag⁺] = 1.10x10⁻²³ M

Explanation:

The equilibrium that takes place is:

Ag⁺ + 2CN⁻  ↔ [Ag(CN)₂]⁻

Kf = 1.0x10²¹ =

We calculate the moles of each reagent:

Ag⁺ ⇒ 90.0 L * 0.21 M = 18.9 moles Ag⁺

CN⁻ ⇒ 90.0 L * 5.0 M = 450 moles CN⁻

Because Ag⁺ is the limiting reactant, we use that value to calculate the moles of [Ag(CN)₂]⁻ formed.

We assume 18.9 moles of [Ag(CN)₂]⁻ are formed, because Kf is quite large.

The moles of CN⁻ that reacted are:

18.9 molAg⁺ * 2molCN⁻/1molAg⁺ = 37.8 moles CN⁻

So the remaining CN⁻ moles in solution are 450-37.8 = 412.2 mol CN⁻

The final volume is 99.0 L, so now we recalculate the molar concentrations of CN⁻ and the complex:

412.2 mol CN⁻ / 99L = 4.16 M

18.9 mol [Ag(CN)₂]⁻ / 99L = 0.19 M

Finally we put the data for the complex and [CN⁻] in the expression of Kf and solve for [Ag⁺]:

= 1.0x10²¹

= 1.0x10²¹

[Ag⁺] = 1.10x10⁻²³ M

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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