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Sagot :
To identify the spectator ions in the given reaction, we need to first understand which ions are present in the reactants and the products and then determine which ions remain unchanged throughout the reaction.
Given reaction:
[tex]\[ 2 H^{+} + CrO_{4}^{2-} + Ba^{2+} + 2 OH^{-} \rightarrow Ba^{2+} + CrO_{4}^{2-} + 2 H_{2}O \][/tex]
### Step-by-Step Solution:
1. List the ions on each side of the reaction:
- Reactants: [tex]\( 2 H^{+} \)[/tex], [tex]\( CrO_{4}^{2-} \)[/tex], [tex]\( Ba^{2+} \)[/tex], [tex]\( 2 OH^{-} \)[/tex]
- Products: [tex]\( Ba^{2+} \)[/tex], [tex]\( CrO_{4}^{2-} \)[/tex], [tex]\( 2 H_{2}O \)[/tex]
2. Identify ions that remain unchanged:
- An ion is considered a spectator ion if it appears in the same form on both sides of the equation.
- [tex]\( CrO_{4}^{2-} \)[/tex] is present on both the reactant and product sides without any change.
- [tex]\( Ba^{2+} \)[/tex] is also present on both the reactant and product sides without any change.
3. Determine the ions that react or form new products:
- [tex]\( H^{+} \)[/tex] combines with [tex]\( OH^{-} \)[/tex] to form water ([tex]\( H_{2}O \)[/tex]).
- [tex]\( OH^{-} \)[/tex] reacts with [tex]\( H^{+} \)[/tex] to form water ([tex]\( H_{2}O \)[/tex]).
Hence, [tex]\( H^{+} \)[/tex], [tex]\( OH^{-} \)[/tex], and [tex]\( H_{2}O \)[/tex] are involved in the chemical reaction and do not remain unchanged.
4. Spectator ions:
- The ions that do not change and remain the same on both sides of the equation are [tex]\( CrO_{4}^{2-} \)[/tex] and [tex]\( Ba^{2+} \)[/tex].
Therefore, the spectator ions in this reaction are:
[tex]\[ CrO_{4}^{2-} \][/tex] and [tex]\[ Ba^{2+} \][/tex]
So the answer is:
- [tex]\( CrO_{4}^{2-} \)[/tex]
- [tex]\( Ba^{2+} \)[/tex]
These are the ions that simply "watch" the reaction occur without getting chemically involved.
Given reaction:
[tex]\[ 2 H^{+} + CrO_{4}^{2-} + Ba^{2+} + 2 OH^{-} \rightarrow Ba^{2+} + CrO_{4}^{2-} + 2 H_{2}O \][/tex]
### Step-by-Step Solution:
1. List the ions on each side of the reaction:
- Reactants: [tex]\( 2 H^{+} \)[/tex], [tex]\( CrO_{4}^{2-} \)[/tex], [tex]\( Ba^{2+} \)[/tex], [tex]\( 2 OH^{-} \)[/tex]
- Products: [tex]\( Ba^{2+} \)[/tex], [tex]\( CrO_{4}^{2-} \)[/tex], [tex]\( 2 H_{2}O \)[/tex]
2. Identify ions that remain unchanged:
- An ion is considered a spectator ion if it appears in the same form on both sides of the equation.
- [tex]\( CrO_{4}^{2-} \)[/tex] is present on both the reactant and product sides without any change.
- [tex]\( Ba^{2+} \)[/tex] is also present on both the reactant and product sides without any change.
3. Determine the ions that react or form new products:
- [tex]\( H^{+} \)[/tex] combines with [tex]\( OH^{-} \)[/tex] to form water ([tex]\( H_{2}O \)[/tex]).
- [tex]\( OH^{-} \)[/tex] reacts with [tex]\( H^{+} \)[/tex] to form water ([tex]\( H_{2}O \)[/tex]).
Hence, [tex]\( H^{+} \)[/tex], [tex]\( OH^{-} \)[/tex], and [tex]\( H_{2}O \)[/tex] are involved in the chemical reaction and do not remain unchanged.
4. Spectator ions:
- The ions that do not change and remain the same on both sides of the equation are [tex]\( CrO_{4}^{2-} \)[/tex] and [tex]\( Ba^{2+} \)[/tex].
Therefore, the spectator ions in this reaction are:
[tex]\[ CrO_{4}^{2-} \][/tex] and [tex]\[ Ba^{2+} \][/tex]
So the answer is:
- [tex]\( CrO_{4}^{2-} \)[/tex]
- [tex]\( Ba^{2+} \)[/tex]
These are the ions that simply "watch" the reaction occur without getting chemically involved.
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