Question

10mL of 1M silver nitrate is combined with 25mL of 0.1M sodium chromate. The Ksp for silver chromate (Ag2CrO4) is 1.1x10-12.
Part A. What is the concentration of sodium ion after the two solutions are combined?
Part B. What is the reaction quotient (Q) of the silver chromate & what is the mass of the silver chromate precipitate in grams?

Answer 1

a) Consider the ionization of sodium chromate, Na₂CrO₄ as below.

Na₂CrO₄ (aq) ---------> 2 Na⁺ (aq) + CrO₄^2- (aq)

As per the stoichiometric equation,

1 mole Na₂CrO₄ = 2 moles Na⁺

Moles of Na₂CrO₄ = (volume of Na₂CrO₄ in L)*(concentration of Na₂CrO₄ in mol/L) = (25 mL)*(1 L/1000 mL)*(0.1 mol/L) [1 M = 1mol/L] = 0.0025 mole.

Therefore, mole(s) of Na⁺ = 2*mole(s) Na₂CrO₄ = (0.0025 mole)*2 = 0.0050 mole.

Total volume of solution after mixing = (10 + 25) mL = 35 mL = (35 mL)*(1 L/1000 mL) = 0.035 L.

Concentration of Na⁺ after mixing = (mole Na⁺)/(volume of solution) = (0.0050 mole)/(0.035 L) = 0.1428 mol/L = 0.1428 M (ans).

b) As per the stoichiometric equation,

2 mole Na⁺ = 1 mole CrO₄^2-

Therefore, concentration of CrO₄^2- in the final solution = ½*(concentration Na⁺) = ½*(0.1428 M) = 0.0714 M.

Use the dilution equation to find out the concentration of Ag⁺ in the final solution.

M₁*V₁ = M₂*V₂

===> (10 mL)*(1 M) = (35 mL)*M₂

===> M₂ = (10*1)/(35) M = 0.2857 M

Write down the dissociation for silver chromate, Ag₂CrO₄.

Ag₂CrO₄ (s) <=====> 2 Ag⁺ (aq) + CrO₄^2- (aq)

We define the reaction quotient as

Q = [Ag⁺]²[CrO₄^2-] = (0.2857)²*(0.0714) = 5.8279*10^-3 ≈ 5.83*10^-3 (ans).

Clearly, CrO₄^2- is the limiting reactant and the concentration of Ag₂CrO₄ precipitated = 0.0714 M.

Mole(s) of Ag₂CrO₄ precipitated = (0.0714 mol/L)*(35 mL)*(1 L/1000 mL) = 0.002499 mole.

Molar mass of Ag₂CrO₄ = 331.73 g/mol.

Mass of Ag₂CrO₄ precipitated = (0.002499 mole)*(331.73 g/mol) = 0.82899 g ≈ 0.83 g (ans).