for the following reaction in aqueous solution, identify the species that will be spectator ions. select all that apply. sr(oh)₂ + k₂s → 2koh + srs choose one or more: a. k₂s b. koh c. s²⁻ d. sr(oh)₂ e. k⁺ f. oh⁻ g. sr²⁺ h. srs

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# Brief Explanations:
To identify spectator ions, we first determine the solubility of each compound and their dissociation in aqueous solution.
- $ \text{Sr(OH)}_2 $: Slightly soluble, but in the reaction context, let's consider dissociation (though it's a weak electrolyte, for the purpose of spectator ions, we analyze ions).
- $ \text{K}_2\text{S} $: Soluble, dissociates into $ 2\text{K}^+ + \text{S}^{2-} $.
- $ \text{KOH} $: Soluble, dissociates into $ \text{K}^+ + \text{OH}^- $.
- $ \text{SrS} $: Insoluble (precipitate), so it does not dissociate.

Now, write the total ionic equation:
$ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-}
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Spectator ions are those that appear unchanged (same charge and state) on both sides. Here, $ \text{K}^+ $ and $ \text{OH}^- $ are present on both sides (reactants and products) in the same form. Wait, no—wait, let's recheck:
Wait, $ \text{Sr(OH)}_2 $ may be considered as dissociating (even if slightly) into $ \text{Sr}^{2+} + 2\text{OH}^- $, $ \text{K}_2\text{S} $ dissociates into $ 2\text{K}^+ + \text{S}^{2-} $, $ \text{KOH} $ dissociates into $ \text{K}^+ + \text{OH}^- $, and $ \text{SrS} $ is a solid.

So the total ionic equation:
$ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-}
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Now, cancel the ions that are the same on both sides: $ 2\text{K}^+ $ and $ 2\text{OH}^- $ are present on both sides? Wait, no—wait, the products are $ 2\text{KOH} $, which is soluble, so $ 2\text{K}^+ + 2\text{OH}^- $, and the reactants have $ 2\text{K}^+ $ (from $ \text{K}_2\text{S} $) and $ 2\text{OH}^- $ (from $ \text{Sr(OH)}_2 $)? Wait, no: $ \text{Sr(OH)}_2 $ is a reactant—if it's a strong base, it dissociates, but $ \text{Sr(OH)}_2 $ is slightly soluble, but in the reaction, maybe we consider it as dissociating for the sake of ionic equation. Wait, maybe I made a mistake. Let's correct:

Reactants:
- $ \text{Sr(OH)}_2 $: Let's assume it dissociates into $ \text{Sr}^{2+} + 2\text{OH}^- $ (even if slightly, for ionic equation).
- $ \text{K}_2\text{S} $: Dissociates into $ 2\text{K}^+ + \text{S}^{2-} $.

Products:
- $ 2\text{KOH} $: Dissociates into $ 2\text{K}^+ + 2\text{OH}^- $.
- $ \text{SrS} $: Solid, so no dissociation.

So total ionic equation:
$ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-}
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Now, the ions that are the same on both sides are $ \text{K}^+ $ and $ \text{OH}^- $? Wait, no—wait, $ \text{OH}^- $ is from $ \text{Sr(OH)}_2 $ (reactant) and $ \text{KOH} $ (product). So $ \text{OH}^- $ is a spectator? And $ \text{K}^+ $ is from $ \text{K}_2\text{S} $ (reactant) and $ \text{KOH} $ (product), so $ \text{K}^+ $ is a spectator. Wait, but let's check the options:

Options:
A. $ \text{K}_2\text{S} $: Compound, not an ion.
B. $ \text{KOH} $: Compound, not an ion.
C. $ \text{S}^{2-} $: Reactant ion, becomes part of solid $ \text{SrS} $, so not a spectator.
D. $ \text{Sr(OH)}_2 $: Compound, not an ion.
E. $ \text{K}^+ $: Spectator (same on both sides).
F. $ \text{OH}^- $: Spectator (same on both sides).
G. $ \text{Sr}^{2+} $: Becomes part of solid $ \text{SrS} $, so not a spectator.
H. $ \text{SrS} $: Solid, not an ion.

Wait, but earlier I thought $ \text{OH}^- $ is from $ \text{Sr(OH)}_2 $ (reactant) and $ \text{KOH} $ (product). So $ \text{OH}^- $ is present on both sides. Similarly, $ \text{K}^+ $ is present on both sides (from $ \text{K}_2\text{S} $ and $ \text{KOH} $). So spectator ions are $ \text{K}^+ $ (E) and $ \text{OH}^- $ (F)? Wait, but let's re-express the reaction:

The net ionic equation is $ \text{Sr}^{2+} + \text{S}^{2-}
ightarrow \text{SrS}(s) $, because $ \text{K}^+ $ and $ \text{OH}^- $ are canceled. So $ \text{K}^+ $ and $ \text{OH}^- $ are spectator ions.

Now, check the options:
E is $ \text{K}^+ $, F is $ \text{OH}^- $. Wait, but the options also include A, B, D, H as compounds, not ions. So the spectator ions are $ \text{K}^+ $ (E) and $ \text{OH}^- $ (F)? Wait, but let's check the original reaction again.

Wait, maybe I made a mistake in the dissociation of $ \text{Sr(OH)}_2 $. If $ \text{Sr(OH)}_2 $ is a solid (insoluble) and does not dissociate, then the ionic equation would be different. Let's consider:

If $ \text{Sr(OH)}_2 $ is a solid (undissociated) and $ \text{K}_2\text{S} $ is soluble (dissociated into $ 2\text{K}^+ + \text{S}^{2-} $), $ \text{KOH} $ is soluble (dissociated into $ \text{K}^+ + \text{OH}^- $), and $ \text{SrS} $ is a solid.

Then the reaction is:
$ \text{Sr(OH)}_2(s) + 2\text{K}^+ + \text{S}^{2-}(aq)
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

In this case, $ 2\text{K}^+ $ is present on both sides (reactants and products), so $ \text{K}^+ $ is a spectator ion. $ \text{OH}^- $ is a product, and $ \text{S}^{2-} $ is a reactant, $ \text{Sr(OH)}_2 $ is a solid, $ \text{SrS} $ is a solid. So here, $ \text{K}^+ $ is a spectator. But $ \text{OH}^- $ is a product, not a reactant, so it's not a spectator (since it wasn't in the reactants). Wait, this is confusing. Let's use the definition: spectator ions are ions that are present in the reaction mixture but do not participate in the actual chemical reaction (i.e., they appear in the same form on both sides of the ionic equation).

So to determine, we need to write the ionic equation with correct dissociation:

1. Identify soluble and insoluble compounds:
   - $ \text{Sr(OH)}_2 $: Slightly soluble (let's assume it's a solid, undissociated).
   - $ \text{K}_2\text{S} $: Soluble (dissociates into $ 2\text{K}^+ + \text{S}^{2-} $).
   - $ \text{KOH} $: Soluble (dissociates into $ \text{K}^+ + \text{OH}^- $).
   - $ \text{SrS} $: Insoluble (solid, undissociated).

2. Write the ionic equation:
   Reactants: $ \text{Sr(OH)}_2(s) + 2\text{K}^+(aq) + \text{S}^{2-}(aq) $
   Products: $ 2\text{K}^+(aq) + 2\text{OH}^-(aq) + \text{SrS}(s) $

3. Now, check which ions are the same on both sides: $ 2\text{K}^+ $ is present on both sides (reactants and products) in aqueous solution. So $ \text{K}^+ $ is a spectator ion.

But what about $ \text{OH}^- $? In the reactants, $ \text{Sr(OH)}_2 $ is a solid, so $ \text{OH}^- $ is not present in the reactants as ions (if $ \text{Sr(OH)}_2 $ is undissociated). So $ \text{OH}^- $ is a product, not a reactant ion. Therefore, $ \text{OH}^- $ is not a spectator (since it wasn't in the reactants). Wait, this is a key point. If $ \text{Sr(OH)}_2 $ is undissociated (solid), then $ \text{OH}^- $ comes from the reaction of $ \text{Sr(OH)}_2 $ with $ \text{K}_2\text{S} $, but actually, the reaction is a double displacement: $ \text{Sr(OH)}_2 + \text{K}_2\text{S}
ightarrow 2\text{KOH} + \text{SrS} $.

In double displacement, the ions exchange: $ \text{Sr}^{2+} $ (from $ \text{Sr(OH)}_2 $) and $ \text{S}^{2-} $ (from $ \text{K}_2\text{S} $) form $ \text{SrS} $ (solid), and $ \text{K}^+ $ (from $ \text{K}_2\text{S} $) and $ \text{OH}^- $ (from $ \text{Sr(OH)}_2 $) form $ \text{KOH} $ (soluble).

So if $ \text{Sr(OH)}_2 $ dissociates into $ \text{Sr}^{2+} + 2\text{OH}^- $ (even slightly), and $ \text{K}_2\text{S} $ dissociates into $ 2\text{K}^+ + \text{S}^{2-} $, then the ionic equation is:

$ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-}
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Here, $ \text{K}^+ $ and $ \text{OH}^- $ are present on both sides (reactants and products) in the same form (aqueous ions), so they are spectator ions. $ \text{Sr}^{2+} $ and $ \text{S}^{2-} $ form the solid $ \text{SrS} $, so they are not spectators.

Therefore, the spectator ions are $ \text{K}^+ $ (E) and $ \text{OH}^- $ (F). Wait, but the options also include A, B, D, H as compounds. The question says "species that will be spectator ions"—ions, so we need to look at the ion options: E ($ \text{K}^+ $) and F ($ \text{OH}^- $)? Wait, but let's check the original options again:

Options:
A. $ \text{K}_2\text{S} $ (compound, not ion)
B. $ \text{KOH} $ (compound, not ion)
C. $ \text{S}^{2-} $ (ion, reacts to form solid)
D. $ \text{Sr(OH)}_2 $ (compound, not ion)
E. $ \text{K}^+ $ (ion, spectator)
F. $ \text{OH}^- $ (ion, spectator)
G. $ \text{Sr}^{2+} $ (ion, reacts to form solid)
H. $ \text{SrS} $ (compound, not ion)

So the spectator ions are $ \text{K}^+ $ (E) and $ \text{OH}^- $ (F). Wait, but maybe I was wrong about $ \text{OH}^- $. Let's confirm with the ionic equation:

If $ \text{Sr(OH)}_2 $ dissociates into $ \text{Sr}^{2+} + 2\text{OH}^- $ (reactants), and $ \text{K}_2\text{S} $ dissociates into $ 2\text{K}^+ + \text{S}^{2-} $ (reactants), then products are $ 2\text{KOH} $ (dissociates into $ 2\text{K}^+ + 2\text{OH}^- $) and $ \text{SrS} $ (solid). So the ionic equation is:

$ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-}
ightarrow 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Here, $ \text{K}^+ $ and $ \text{OH}^- $ are identical on both sides (same charge, same state: aqueous), so they are spectator ions. Thus, E ($ \text{K}^+ $) and F ($ \text{OH}^- $) are spectator ions.

Wait, but the initial options—maybe the question considers $ \text{Sr(OH)}_2 $ as dissociated, so $ \text{OH}^- $ is a spectator. So the correct spectator ions are $ \text{K}^+ $ (E) and $ \text{OH}^- $ (F). But let's check the answer options again. The user is to select all that apply. So E and F? Wait, but maybe I made a mistake. Let's check a reference: Spectator ions are ions that do not participate in the reaction, i.e., they are present in the same form on both sides. In this reaction, the net ionic equation is $ \text{Sr}^{2+} + \text{S}^{2-}
ightarrow \text{SrS}(s) $, so the spectator ions are $ \text{K}^+ $ and $ \text{OH}^- $, which are canceled out. So yes, E ($ \text{K}^+ $) and F ($ \text{OH}^- $) are spectator ions.

Wait, but the options also include A, B, D, H as compounds. The question says "species that will be spectator ions"—ions, so only E and F? But let's check the original reaction again. Wait, $ \text{Sr(OH)}_2 $ is a base, $ \text{K}_2\text{S} $ is a salt. The reaction is a double displacement, forming $ \text{KOH} $ (soluble) and $ \text{SrS} $ (insoluble). So the ionic equation:

Reactants: $ \text{Sr}^{2+} + 2\text{OH}^- + 2\text{K}^+ + \text{S}^{2-} $
Products: $ 2\text{K}^+ + 2\text{OH}^- + \text{SrS}(s) $

Thus, $ \text{K}^+ $ and $ \text{OH}^- $ are spectators. So the correct options are E ($ \text{K}^+ $) and F ($ \text{OH}^- $). Wait, but the user's options: E is $ \text{K}^+ $, F is $ \text{OH}^- $. So those are the spectator ions.
# Answer:
E. $ \text{K}^+ $, F. $ \text{OH}^- $

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