QuestionHomework Chapter 4A Question 9 of 11 (1 point) I Question Attempt: 1 of Unlimited $\checkmark 1$ $\checkmark 2$ $\checkmark 3$ 4 $\checkmark 5$ $\checkmark 7$ 8 9 10
Write the net ionic equation for the reaction between $\mathrm{FeCl}_{3}(\mathrm{aq})$ and $\mathrm{KOH}(\mathrm{aq})$ . Be sure to include states of matter in your answer. Note: Reference the Solubility of ionic compounds in water table for additional information.

Studdy Solution

STEP 1

Assumptions
1. We are given two aqueous solutions: $\mathrm{FeCl}_{3}(\mathrm{aq})$ and $\mathrm{KOH}(\mathrm{aq})$ .
2. The reaction between $\mathrm{FeCl}_{3}$ and $\mathrm{KOH}$ will produce a precipitate.
3. We need to write the net ionic equation, which includes only the species that participate in the reaction.
4. Solubility rules will help determine which compounds are soluble and which form precipitates.

STEP 2

Write the balanced molecular equation for the reaction between $\mathrm{FeCl}_{3}(\mathrm{aq})$ and $\mathrm{KOH}(\mathrm{aq})$ .
$\mathrm{FeCl}_{3}(\mathrm{aq}) + 3\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{Fe(OH)}_{3}(\mathrm{s}) + 3\mathrm{KCl}(\mathrm{aq})$

STEP 3

Identify the states of matter for each compound in the equation:
- $\mathrm{FeCl}_{3}(\mathrm{aq})$ : aqueous - $\mathrm{KOH}(\mathrm{aq})$ : aqueous - $\mathrm{Fe(OH)}_{3}(\mathrm{s})$ : solid (precipitate) - $\mathrm{KCl}(\mathrm{aq})$ : aqueous

STEP 4

Write the complete ionic equation by dissociating all strong electrolytes (aqueous compounds) into their ions:
$\mathrm{Fe}^{3+}(\mathrm{aq}) + 3\mathrm{Cl}^{-}(\mathrm{aq}) + 3\mathrm{K}^{+}(\mathrm{aq}) + 3\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Fe(OH)}_{3}(\mathrm{s}) + 3\mathrm{K}^{+}(\mathrm{aq}) + 3\mathrm{Cl}^{-}(\mathrm{aq})$

STEP 5

Identify and cancel the spectator ions, which appear on both sides of the equation:
- Spectator ions: $\mathrm{K}^{+}(\mathrm{aq})$ and $\mathrm{Cl}^{-}(\mathrm{aq})$

STEP 6

Write the net ionic equation by removing the spectator ions:
$\mathrm{Fe}^{3+}(\mathrm{aq}) + 3\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Fe(OH)}_{3}(\mathrm{s})$
This is the net ionic equation for the reaction between $\mathrm{FeCl}_{3}(\mathrm{aq})$ and $\mathrm{KOH}(\mathrm{aq})$ .

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