since Ag and Cl are in equilibrium with AgCl, find Ksp for agcl from the concentrations. Write the expression for ksp AgCl and determine its value

When Ag⁺ and Cl⁻ are in equilibrium with AgCl, the expression for Ksp is

And its value is 1.77 × 10⁻¹⁰.

Let's consider the equation for the solution of AgCl.

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

The molar solubility of AgCl (S) at 25 °C is 1.33 × 10⁻⁵ M. We can use this information to calculate the solubility product (Ksp) through an ICE chart.

       AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                          0             0

C                       +S           +S

E                        S              S

The solubility product (Ksp) for AgCl is:

When Ag⁺ and Cl⁻ are in equilibrium with AgCl, the expression for Ksp is

And its value is 1.77 × 10⁻¹⁰.

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1.69 ×10^-10

Explanation:

Given that the equation for the dissolution of AgCl in water is;

AgCl(s) ⇄Ag^+(aq) + Cl^+(aq)

Also, silver ion and chloride ion are in equilibrium with the undissociated AgCl hence we can write;

Ksp= [Ag+][Cl-]/ [AgCl]

Since the activity of the pure sold is 1, we now have;Ksp= [Ag+][Cl-]

If we know the solubility of AgCl in pure water to be 1.3 x 10^-5 M, from standard tables, and [Ag+]=[Cl-]= 1.3 x 10^-5 M = x

Then;

Ksp= x^2

Ksp= (1.3 x 10^-5)^2

Ksp= 1.69 ×10^-10

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.