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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.