Answer:
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
When Ag⁺ and Cl⁻ are in equilibrium with AgCl, the expression for Ksp is
[tex]Ksp = [Ag^{+} ][Cl^{+} ][/tex]
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:
[tex]Ksp = [Ag^{+} ][Cl^{+} ] = S.S = S^{2} = (1.33 \times 10^{-5} )^{2} = 1.77 \times 10^{-10}[/tex]
When Ag⁺ and Cl⁻ are in equilibrium with AgCl, the expression for Ksp is
[tex]Ksp = [Ag^{+} ][Cl^{+} ][/tex]
And its value is 1.77 × 10⁻¹⁰.
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