Question

The ksp of agcl is 1.6 x 10-10. what is the solubility of agcl in 0.010 m fecl3? give your answer using scientific notation (1.23e-4) and to 2 significant figures (i.e., one decimal place). [blank1]

Answer 1

Ksp of AgCl= 1.6×10⁻¹⁰

AgCl=Ag⁺ +Cl⁻

Ksp=[Ag⁺][Cl⁻]

Assume [Ag⁺]=[Cl⁻]=x

Ksp=x²

1.6×10⁻¹⁰=x²

x=0.000012

In FeCl₃:

FeCl₃------>Fe⁺³+ 3Cl⁻

as there is 0.010 M FeCl₃

So there will be ,

[Cl⁻]= 0.030

So

[Ag⁺]=Ksp/[Cl⁻]

=1.6×10⁻¹⁰/0.030

=5.3×10⁻⁹

so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.

Answer 2

Answer:

5.3 × 10⁻⁹ M

Explanation:

FeCl₃ is a strong electrolyte that ionizes according to the following equation.

FeCl₃(aq) → Fe³⁺(aq) + 3 Cl⁻(aq)

The concentration of Cl⁻ will be 3 × [FeCl₃(aq)] = 3 × 0.010 M = 0.030 M

We can find the solubility (S) of AgCl using an ICE Chart.

        AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                           0            0.030

C                         +S            +S

E                          S           0.030 + S

The solubility product (Ksp) is:

Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S . (0.030 + S)

In the term (0.030 + S), S <<< 0.030, so we can neglect S to simplify calculations.

1.6 × 10⁻¹⁰ = 0.030 S

S = 5.3 × 10⁻⁹ M