Question

How many grams of KSCN are required to just completely dissolve 1.44 g AgCl(s) in 0.10 L water? [Kf for Ag(SCN)4^3– is 1.2 ×1010; Ksp(AgCl) = 1.6 ×10^10]

Answer 1

1. As per the given data, Ksp = 1.6 x 10¹⁰

Ag⁺ + Cl^- AgCl(s)

ksp = [Ag⁺][Cl^-] = 1.6 x 10¹⁰

x. x = 1.6 x 10¹⁰

Therefore solubility x = 1.264 x 10⁵ mol/L

amount of AgCl = 1.44g (m.mass = 143.32 g/mol)

[AgCl] = 0.01 mol/0.1 L= 0.1004 M

therefore moles of AgCl = 1.44/143.32 = 0.01004 moles

For 0.1004M of AgCl, solubility would be,

= 1.264 x 10⁵ mol/1L X 0.1004 mol/1L

= 0.1269 x 10⁵ mol/L

Now,

Ag⁺ (aq) + 4 SCN⁻ (aq) Ag(SCN)₄³⁻ (aq) Kf = 1.2×10¹⁰

Since the formation constant is very large, we can say that the reaction has gone to completion

1 mole of Ag⁺ requires 4 moles of SCN⁻

Therefore, 0.1269 x 10⁵ mol/L of Ag⁺ would require 4 x 0.1269 x 10⁵ mol/L of KSCN = 0.5076 x 10⁵ mol/L of KSCN

= 49.32 x 10⁵ g/L or 4.932 x 10⁵ g/0.1L of KSCN

2. assuming Ksp = 1.6 x 10^-10

As per the given data

Ag⁺ + Cl^- AgCl(s)

ksp = [Ag⁺][Cl^-] = 1.6 x 10^-10

x. x = 1.6 x 10^-10

Therefore solubility x = 1.264 x 10^-5 mol/L

amount of AgCl = 1.44g (m.mass = 143.32 g/mol)

[AgCl] = 0.01 mol/0.1 L= 0.1004 M

therefore moles of AgCl = 1.44/143.32 = 0.01004 moles

For 0.1004M of AgCl, solubility would be,

= 1.264 x 10^-5 mol/1L X 0.1004 mol/1L

= 0.1269 x 10^-5 mol/L

Now,

Ag⁺ (aq) + 4 SCN⁻ (aq) Ag(SCN)₄³⁻ (aq) Kf = 1.2×10¹⁰

Since the formation constant is very large, we can say that the reaction has gone to completion

1 mole of Ag⁺ requires 4 moles of SCN⁻

Therefore, 0.1269 x 10^-5 mol/L of Ag⁺ would require 4 x 0.1269 x 10^-5 mol/L of KSCN = 0.5076 x 10^-5 mol/L of KSCN

= 49.32 x 10^-5 g/L or 4.932 x 10^-5 g/0.1L of KSCN