Answer:
Explanation:
1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.
The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:
The ICE (initial, change, equilibrium) table is:
I X 0 0
C -s +s +s
E X - s s s
Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.
Thus, the Ksp equation is:
By substitution:
Rounding to two significant figures:
2. Molar solubility of AgCl(s) in seawater
Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.
The new ICE table will be:
I X 0 0.54
C -s +s +s
E X - s s s + 0.54
The new equation for the Ksp equation will be:
By substitution:
Now you must solve a quadratic equation.
Use the quadratic formula:
[tex]s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}[/tex]
The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer
3. Why is AgCl(s) less soluble in seawater than in distilled water.
AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.
This is known as the common ion effect.
By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.
In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.
The concentration of Ag ion in the AgCl solution has been 1.3 [tex]\rm \times\;10^-^5[/tex] M.
The calculation of the solubility product constant for the reaction can be given by the product of the concentration of the reactants.
The reactants in AgCl has been the Ag ions and the Cl ions.
(a) The solubility product constant:
[tex]\rm k_s_p\;=\;Ag^+\;+\;Cl^-[/tex]
The equilibrium has an equal concentration of Ag and Cl. Thus, let the concentration of both Ag and Cl be x.
1.8 [tex]\rm \times\;10^-^1^0[/tex] = x [tex]\times[/tex] x
1.8 [tex]\rm \times\;10^-^1^0[/tex] = [tex]\rm x^2[/tex]
x = [tex]\rm \sqrt{1.8\;\times\;10^-^1^0}[/tex]
x = 1.3 [tex]\rm \times\;10^-^5[/tex] M.
The concentration of Ag ion in the AgCl solution has been 1.3 [tex]\rm \times\;10^-^5[/tex] M.
(b) When the initial concentration of Cl ion has been 0.54 M.
The ICE table will be:
AgCl Ag Cl
I x 0 0.54
C -b +b +b
E x - b b 0.54 + b
The molar solubility can be given by:
[tex]\rm k_s_p\;=\;Ag^+\;+\;Cl^-[/tex]
[tex]\rm k_s_p[/tex] = b [tex]\times[/tex] 0.54 + b
[tex]\rm k_s_p[/tex] = 0.54 + [tex]\rm b^2[/tex]
1.8 [tex]\rm \times\;10^-^1^0[/tex] = 0.54 + [tex]\rm b^2[/tex]
b = 3.3 [tex]\rm \times\;10^-^1^0[/tex] M.
The concentration of Cl ion in the seawater has been 3.3 [tex]\rm \times\;10^-^1^0[/tex] M.
(c) The Cl ion has been less soluble in seawater due to the common ion shift. The presence of Cl ion in the seawater shifts the equilibrium and thereby the solubility of the AgCl has been affected.
For more information about the solubility constant, refer to the link:
https://brainly.com/question/13949246
