Answer:
I. Solution will form a precipitate.
II. Solution will not form a precipitate.
Explanation:
I. 0,00090g of Na₂CrO₄ are:
0,00090g×[tex]\frac{1mol}{162,0g}[/tex]= 5,56x10⁻⁶ moles of Na₂CrO₄≡ moles of CrO₄²⁻.
As volume of solution is 225mL:
5,56x10⁻⁶ moles of CrO₄²⁻ / 0,225L = 2,47x10⁻⁵ M of CrO₄²⁻.
The molar concentration of Ag⁺ is 0,00025M.
Ag⁺ with CrO₄²⁻ ions produce:
Ag₂CrO₄(s) ⇄ 2Ag⁺(aq) + CrO₄²⁻(aq)
Where kps is:
kps = 1,1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]
Replacing:
1,1x10⁻¹² = [0,00025M]² [2,47x10⁻⁵M]
1,1x10⁻¹² = 1,5x10⁻¹²
As the product of [Ag⁺]² [CrO₄²⁻] is higher than kps, the ions will decrease it concentration producing the precipitate.
II. Total volume of the solution is 0,300L, that means concentration of MgCl₂ and NaF will be:
0,0015M MgCl₂×[tex]\frac{0,100L}{0,300L}[/tex]= 0,0005M MgCl₂≡ Mg²⁺
0,012M NaF×[tex]\frac{0,200L}{0,300L}[/tex]= 0,008M NaF ≡ F⁻
F⁻ and Mg²⁺ ions produce:
MgF₂(s) ⇄ 2F⁻(aq) + Mg²⁺(aq)
Where kps is defined as:
kps = 3,7x10⁻⁸ = [F⁻]² [Mg²⁺]
Replacing:
3,7x10⁻⁸ = [0,008M]² [0,0005M]
3,7x10⁻⁸ = 3,2x10⁻⁸
As kps is higher than [F⁻]² [Mg²⁺], the ions will not form the precipitate.
I hope it helps!
