In which solution(s) will a precipitate form? i. 0.00090 g of na2cro4 (molar mass = 162.0 g/mol) is added to 225 ml of 0.00025 m agno3. ksp ag2cro4 =1.1 x 10-12 ii. 0.100l of 0.0015 m mgcl2 and 0.200l of 0.012 m naf. ksp mgf2 = 3.7 x 10-8

I. Solution will form a precipitate.

II. Solution will not form a precipitate.

Explanation:

I. 0,00090g of Na₂CrO₄ are:

0,00090g×= 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×= 0,0005M MgCl₂≡ Mg²⁺

0,012M NaF×= 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!

b. is the answer hoped i        

c. orbital shape i hope this is correct