contestada

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]

Respuesta :

dsdrajlin

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

pstnonsonjoku

The solubility of AgCl in 0.01 M FeCl3 is 5.3 x 10-9 M.

Let the solubility of AgCl in FeCl3 be S

The concentration of Cl^- = 3(0.010) = 0.030 M

To solve this question, we have to set up the ICE table as shown;

AgCl(s) ⇄                 Ag^+(aq)  + Cl^-(aq)

1.6 x 10-10                    0                  0    

-S                                 +S                 +S

1.6 x 10-10 - S                S                 0.030  + S

Since Ksp = [Ag^+] [Cl^]

Ksp = S(0.030 + S)

Ksp value is  very small

S value can be ignored, with respect to 0.030 M

1.6 x 10-10 = S × 0.030

S = 5.3 x 10-9 M

Learn more: https://brainly.com/question/10038290

Otras preguntas