A solution of iron(III) chloride was poured into a potassium phosphate solution. Would you expect a precipitate to form if 426.0 mL of the iron(III) chloride solution (2.50 x 10-11 M) was mixed with 410.0 mL of a 1.75 x 10-9 M potassium phosphate solution?

Check Explanation

Explanation:

The reaction between Iron III chloridr and potassium phosphate is represented by

FeCl₃ (aq) + K₃PO₄ (aq) → FePO₄ (s) + 3KCl (aq)

The expected precipitate is FePO₄.

Note that

Number of moles = (Concentration in mol/L) × (Volume in L)

For FeCl₃

Concentration of FeCl₃ = (2.50 × 10⁻¹¹) M

Volume in L = (426/1000) = 0.426 L

Number of moles of FeCl₃ = (2.50 × 10⁻¹¹) × 0.426 = (1.065 × 10⁻¹¹) mole

For K₃PO₄

Concentration of K₃PO₄ = (1.75 × 10⁻⁹) M

Volume in L = (410/1000) = 0.410 L

Number of moles of K₃PO₄ = (1.75 × 10⁻⁹) × 0.410 = (7.175 × 10⁻¹⁰) mole

FeCl₃ is the limiting reagent as it is the reagent in short supply. It is the reactant that determines the amount of the other reactant that reacts and the amount of products that will be formed.

Since the precipitate expected is FePO₄

1 mole of FeCl₃ gives 1 mole of FePO₄

(1.065 × 10⁻¹¹) mole of FeCl₃ will give (1.065 × 10⁻¹¹) mole of FePO₄

So, this reaction results in the formation of (1.065 × 10⁻¹¹) mole of FePO₄

In terms of mass,

Mass of FePO₄ formed = (Number of moles) × (Molar Mass) = (1.065 × 10⁻¹¹) × (150.82)

= (1.606 × 10⁻⁹) g

Normally, a precipitate would form from this reaction because the resulting Iron III phosphate from the reaction isn't soluble in water, but the amount of FePO₄ precipitate formed, (1.606 × 10⁻⁹) g, is too small to be noticeable in the reaction mixture.

Hence, it would seem like no precipitate is formed for this reaction.

Hope this Helps!!!

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