Much of the earth’s heat is retained in the crust. wind. oceans. mountains.
2 answers:
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Answer:
- [Ag⁺] = 1.3 × 10⁻⁵M
- s = 3.3 × 10⁻¹⁰ M
- Because the common ion effect.
Explanation:
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<u>1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.</u>
The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:
- AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)
The ICE (initial, change, equilibrium) table is:
- AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)
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:
- Ksp = [Ag⁺] [Cl⁻]
- Ksp = s × s
- Ksp = s²
By substitution:
- 1.8 × 10⁻¹⁰ = s²
- s = 1.34 × 10⁻⁵M
Rounding to two significant figures:
- [Ag⁺] = 1.3 × 10⁻⁵M ← answer
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<u>2. Molar solubility of AgCl(s) in seawater</u>
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:
- AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)
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:
- Ksp = [Ag⁺] [Cl⁻]
- Ksp = s × ( s + 0.54)
- Ksp = s² + 0.54s
By substitution:
- 1.8 × 10⁻¹⁰ = s² + 0.54s
- s² + 0.54s - 1.8 × 10⁻¹⁰ = 0
Now you must solve a quadratic equation.
Use the quadratic formula:
The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer
<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>
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.