Solutions of iron(III) chloride and sodium hydroxide were mixed in a series of precipita- tion reactions, as described in this experiment. (a) Name the two possible products in this precipitation reaction and predict their empir- ical formulas. (b) Which product is likely to be insoluble in water and precipitate out as a red solid? (c) What volume ratio of reactants gave the most precipitate (see Table 1)? Explain. Table 1. Test tube. 1 2. 3 6. FeCl,, 0.1 M, mL 5 10 12 15 17 20 24 NaOH, 0.1 M, mL 55 50 48 45 43 40 36 Volume of precipitate, mL 1 10 14 20 4 1
The Correct Answer and Explanation is :
When solutions of iron(III) chloride (FeCl₃) and sodium hydroxide (NaOH) are mixed, a precipitation reaction occurs, leading to the formation of iron(III) hydroxide (Fe(OH)₃) and sodium chloride (NaCl).
(a) Name the two possible products in this precipitation reaction and predict their empirical formulas.
The two products are:
- Iron(III) hydroxide: This compound forms as a reddish-brown precipitate when NaOH is added to FeCl₃. Its empirical formula is Fe(OH)₃.
- Sodium chloride: This is a soluble salt formed in the reaction. Its empirical formula is NaCl.
(b) Which product is likely to be insoluble in water and precipitate out as a red solid?
Iron(III) hydroxide (Fe(OH)₃) is insoluble in water and precipitates out as a reddish-brown solid when NaOH is added to FeCl₃.
(c) What volume ratio of reactants gave the most precipitate (see Table 1)? Explain.
To determine the volume ratio of FeCl₃ to NaOH that produced the most precipitate, we need to analyze the data provided in Table 1:
| Test Tube | FeCl₃ (0.1 M) Volume (mL) | NaOH (0.1 M) Volume (mL) | Volume of Precipitate (mL) |
|---|---|---|---|
| 1 | 5 | 55 | 1 |
| 2 | 10 | 50 | 10 |
| 3 | 12 | 48 | 14 |
| 4 | 15 | 45 | 20 |
| 5 | 17 | 43 | 4 |
| 6 | 20 | 40 | 1 |
| 7 | 24 | 36 | 0 |
From the table, the maximum volume of precipitate (20 mL) was obtained in Test Tube 4, where 15 mL of FeCl₃ solution was mixed with 45 mL of NaOH solution.
Explanation:
The precipitation reaction between FeCl₃ and NaOH is governed by the stoichiometric relationship:
[ \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} ]
This equation indicates that one mole of FeCl₃ reacts with three moles of NaOH.
In Test Tube 4, the molar amounts of FeCl₃ and NaOH are:
- FeCl₃: ( 0.1\,\text{M} \times 0.015\,\text{L} = 0.0015\,\text{mol} )
- NaOH: ( 0.1\,\text{M} \times 0.045\,\text{L} = 0.0045\,\text{mol} )
The molar ratio of FeCl₃ to NaOH is 1:3, which matches the stoichiometric requirement. This optimal ratio ensures that both reactants are completely consumed, leading to the maximum formation of Fe(OH)₃ precipitate.
In other test tubes, deviations from this ratio result in less precipitate formation. For example, in Test Tube 5, the molar ratio is approximately 1:2.53, leading to only 4 mL of precipitate. In Test Tube 7, the molar ratio is about 1:1.5, resulting in no precipitate formation.
Therefore, the volume ratio of FeCl₃ to NaOH that produced the most precipitate is 1:3, corresponding to Test Tube 4.