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Chromate and ferrous ions react in acidic solution to form chromic and ferric ions, respectively. If 0.9422 grams of iron(II) bisulfate dissolved in sulfuric acid solution requires 24.83 mL of sodium chromate solution for complete titration, what is the molarity of the sodium chromate solution? Show the balanced net ionic and complete chemical equations for this reaction.

Respuesta :

dsdrajlin

Answer:

0.1518M

Explanation:

Let's consider the balanced complete chemical equation for this double displacement reaction.

Fe(HSO₄)₂(aq) + Na₂CrO₄(aq) ⇄ 2 NaHSO₄(aq) + FeCrO₄(s)

We can write it as a full ionic equation.

Fe²⁺(aq) +  HSO₄⁻(aq) + 2 Na⁺(aq) + CrO₄²⁻(aq) ⇄ 2 Na⁺(aq) + HSO₄⁻(aq) + FeCrO₄(s)

We can also write the net ionic equation.

Fe²⁺(aq) + CrO₄²⁻(aq) ⇄ FeCrO₄(s)

We know the following relations:

  • The molar mass of Fe(HSO₄)₂ is 249.99 g/mol.
  • 1 mole of Fe(HSO₄)₂ reacts with 1 mole of Na₂CrO₄

Then, for 0.9422 grams of iron(II) bisulfate:

[tex]0.9422gFe(HSO_{4})_{2}.\frac{1molFe(HSO_{4})_{2}}{249.99gFe(HSO_{4})_{2}} .\frac{1molNa_{2}CrO_{4}}{1molFe(HSO_{4})_{2}} =3.769 \times 10^{-3} molNa_{2}CrO_{4}[/tex]

The molarity of the sodium chromate solution is:

[tex]M=\frac{n}{V(L)} =\frac{3.769 \times 10^{-3} mol}{0.02483L} =0.1518M[/tex]