Hydrated ammonium iron (II) sulfate crystals have the formula (NH4)2SO4FeSO4 xH2O. 8.325 g (394.2g/mol) of the salt were dissolved in 250.0 ml of acidified water. A 25.0 mL portion of this solution was titrated with potassium manganate (VII) solution of concentration 0.148 M. A volume of 2.250 mL was required. Calculate the value of x.
The correct answer and explanation is :
To determine the value of ‘x’ in the formula (NH₄)₂SO₄·FeSO₄·xH₂O for hydrated ammonium iron(II) sulfate, we can utilize the results of a potassium permanganate (KMnO₄) titration. Here’s a step-by-step explanation:
1. Understanding the Reaction:
In acidic conditions, KMnO₄ reacts with Fe²⁺ ions according to the balanced equation:
[ \text{MnO}_4^- (aq) + 5 \, \text{Fe}^{2+} (aq) + 8 \, \text{H}^+ (aq) \rightarrow \text{Mn}^{2+} (aq) + 5 \, \text{Fe}^{3+} (aq) + 4 \, \text{H}_2\text{O} (l) ]
This equation indicates that one mole of MnO₄⁻ reacts with five moles of Fe²⁺.
2. Calculating Moles of KMnO₄ Used:
The volume of KMnO₄ solution used in the titration is given as 2.250 mL (0.002250 L) with a concentration of 0.148 M. The moles of KMnO₄ reacted are:
[ \text{Moles of KMnO}_4^- = 0.148 \, \text{mol/L} \times 0.002250 \, \text{L} = 0.000333 \, \text{mol} ]
3. Determining Moles of Fe²⁺ in the Titrated Sample:
Using the stoichiometric ratio from the balanced equation (1:5), the moles of Fe²⁺ in the 25.0 mL sample are:
[ \text{Moles of Fe}^{2+} = 5 \times 0.000333 \, \text{mol} = 0.001665 \, \text{mol} ]
4. Extrapolating to the Total Moles of Fe²⁺ in the Original Solution:
Since the 25.0 mL sample is one-tenth of the total 250.0 mL solution, the total moles of Fe²⁺ in the original solution are:
[ \text{Total moles of Fe}^{2+} = 0.001665 \, \text{mol} \times 10 = 0.01665 \, \text{mol} ]
5. Calculating the Mass of Fe²⁺ in the Original Solution:
The molar mass of Fe is approximately 55.85 g/mol, so the mass of Fe²⁺ is:
[ \text{Mass of Fe}^{2+} = 0.01665 \, \text{mol} \times 55.85 \, \text{g/mol} = 0.930 \, \text{g} ]
6. Determining the Moles of Anhydrous Ammonium Iron(II) Sulfate:
The total mass of the salt used is 8.325 g. The moles of the anhydrous salt (FeSO₄·(NH₄)₂SO₄) are:
[ \text{Moles of salt} = \frac{8.325 \, \text{g}}{394.2 \, \text{g/mol}} = 0.02112 \, \text{mol} ]
7. Calculating the Number of Moles of Water (x):
The difference between the moles of Fe²⁺ and the moles of anhydrous salt corresponds to the moles of water of crystallization:
[ \text{Moles of water} = 0.02112 \, \text{mol} – 0.01665 \, \text{mol} = 0.00447 \, \text{mol} ]
8. Determining the Value of x:
Each formula unit of the hydrated salt contains one Fe²⁺ ion. Therefore, the number of moles of water per mole of salt is:
[ x = \frac{0.00447 \, \text{mol}}{0.02112 \, \text{mol}} \times 100 = 21.2\% ]
This percentage corresponds to approximately 2 water molecules per formula unit, suggesting that x ≈ 2.
Conclusion:
Based on the titration data and calculations, the value of ‘x’ in the formula (NH₄)₂SO₄·FeSO₄·xH₂O is approximately 2, indicating that the hydrated salt contains 2 molecules of water of crystallization per formula unit.