Answer:
A precipitate will form.
[Ag⁺] = 2.8x10⁻⁵M
Explanation:
When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:
Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)
Ksp is defined as:
Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]
<em>Where the concentrations [] are in equilibrium</em>
Reaction quotient, Q, is defined as:
Q = [Ag⁺]² [CrO₄²⁻]
<em>Where the concentrations [] are the actual concentrations</em>
<em />
If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,
The actual concentrations are -Where 500mL is the total volume of the solution-:
[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M
[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M
And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴
As Q > Ksp; a precipitate will form
In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:
[Ag⁺] = 0.06M - 2X
[CrO₄²⁻] = 0.165M - X
<em>Where X is defined as the reaction coordinate</em>
<em />
Replacing in Ksp expression:
1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]
Solving for X:
X = 0.165M → False solution. Produce negative concentrations.
X = 0.0299986M
Replacing, equilibrium concentrations are:
[Ag⁺] = 0.06M - 2(0.0299986M)
[CrO₄²⁻] = 0.165M - 0.0299986M
<h3>[Ag⁺] = 2.8x10⁻⁵M</h3>
[CrO₄²⁻] = 0.135M
