Question

6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) solution required 15.8 mL of a 0.254 M solution of dichromate to reach the equivalence point, what is the molarity of the iron(II) solution?

Answer 1

Answer:

1.047 M

Explanation:

The given reaction:

$6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}$

For dichromate :

Molarity = 0.254 M

Volume = 15.8 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

Thus, moles of dichromate :

$Moles=0.254 \times {15.8\times 10^{-3}}\ moles$

Moles of dichromate = 0.0040132 moles

1 mole of dichromate react with 6 moles of iron(II) solution

Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

Moles of iron(II) solution = 0.02408 moles

Volume = 23 mL = 0.023 L

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Molarity = 0.02408 / 0.023 = 1.047 M