Identify the spectator ion(s) in the following reaction:
Cu(OH)2(s) + 2H+(aq) + 2Cl–(aq) Cu2+(aq) + 2Cl–(aq) + 2H2O(l)
a. Cu2+ and Cl-
b. H+ and Cl–
c. Cu(OH)2
d. Cl–
e. Cu2+
With no arrow it is difficult to know which are the reactants and which are the products. To answer your question:
1. Write the arrow where it belongs and do that EVERY time you write an equation. You CAN'T omit the arrow and know what you're doing.
2. Identify the ions on each side of the arrow that are common. Those are the spectator ions.
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To identify the spectator ions in a chemical reaction, we need to determine which ions remain unchanged and do not participate in the actual chemical reaction.
In this reaction, Cu(OH)2(s) + 2H+(aq) + 2Cl–(aq) Cu2+(aq) + 2Cl–(aq) + 2H2O(l), we can see that the Cu2+ ion and the Cl- ion appear both on the reactant side and the product side of the equation. These ions are not directly involved in the reaction and do not undergo any changes. Therefore, they are considered spectator ions.
So, the correct answer is:
a. Cu2+ and Cl-