"Add the ionic equations for part one and three, compare the results with the ionic equation for part two"
Part 1: NaOH(s) = Na+(aq) +OH-(aq) + heat
Part 2: NaOH(s) + H+(aq) +Cl-(aq) = Na+(aq) + Cl-(aq) + H2O(l) + heat
Part 3: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) = Na+(aq) + Cl-(aq) + H2O(l) + heat
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So I thought if you add part 1 and two, you'd get an equation that looks like this:
NaOH + H + Cl = Na + Cl + H2O + 2 heats?????
Is this what I'm even supposed to do? This is part of an enthalpy lab, if that helps. I'd really, really appreciate help.
I feel like I'm totally wrong and not grasping something somehow.
To compare the results between the ionic equations for parts one and three with the ionic equation for part two, you need to combine the relevant species on both sides of the equations. This will allow you to see the similarities and differences between the three equations.
Let's break down each part and write the respective ionic equations:
Part 1:
NaOH(s) → Na+(aq) + OH-(aq) + heat
This equation represents the dissociation of solid sodium hydroxide (NaOH) into its ions, sodium cation (Na+) and hydroxide anion (OH-). The heat in the equation indicates an exothermic process, meaning it releases heat.
Part 2:
NaOH(s) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) + heat
This equation represents the reaction between solid sodium hydroxide (NaOH), aqueous hydrochloric acid (HCl), and water (H2O). In this reaction, the sodium hydroxide reacts with the hydrochloric acid to form sodium chloride (NaCl) and water. Again, the heat indicates an exothermic process.
Part 3:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) + heat
In this equation, the hydroxide ion (OH-) reacts with the hydrogen ion (H+) from the acid, forming water (H2O). The sodium and chloride ions are spectator ions as they do not participate in the reaction. Like the previous equations, this reaction also releases heat.
Now let's compare the three equations:
- Part 1 only involves the dissociation of NaOH into its respective ions, Na+ and OH-. It does not include any other reactants or products.
- Part 2 uses NaOH, H+, and Cl- to form Na+, Cl-, and H2O. This equation shows the neutralization of an acid with a base.
- Part 3 shows that the hydroxide ion (OH-) reacts with the hydrogen ion (H+) and chloride ion (Cl-) to form water (H2O). This equation is similar to part 2 but explicitly includes the hydroxide ion.
Based on this comparison, you can see that part 2 and part 3 have a similar ionic equation. The only difference is that part 3 explicitly includes the hydroxide ion (OH-) instead of using NaOH as in part 2.
To summarize, the ionic equation for part 2 can be compared with the ionic equation for part 3 to see that they represent similar reactions, but part 3 explicitly includes the hydroxide ion. Part 1, on the other hand, represents the dissociation of solid NaOH into its ions and does not involve any other reactants or products from parts 2 or 3.
To add the ionic equations for parts one and three, we will consider the relevant ions and their charges, and then combine them to form the overall equation.
Part 1: NaOH(s) = Na+(aq) + OH-(aq) + heat
Part 3: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) = Na+(aq) + Cl-(aq) + H2O(l) + heat
To compare with the ionic equation for part two:
Part 2: NaOH(s) + H+(aq) + Cl-(aq) = Na+(aq) + Cl-(aq) + H2O(l) + heat
Now let's compare the three equations step by step:
1. In part 1 and part 2, both equations involve Na+(aq) and Cl-(aq) ions. These ions do not change in any of the reactions.
2. In part 1, NaOH(s) dissociates into Na+(aq) and OH-(aq) ions.
3. In part 2, NaOH(s) also dissociates into Na+(aq) and OH-(aq) ions.
4. In part 2, H+(aq) and Cl-(aq) ions join with the Na+(aq) and OH-(aq) ions to form Na+(aq), Cl-(aq), and H2O(l). This reaction consumes heat.
5. In part 3, the H+(aq) and Cl-(aq) ions also join with Na+(aq) and OH-(aq) ions to form Na+(aq), Cl-(aq), and H2O(l). This reaction also consumes heat.
Comparing the results, we can see that the sum of part 1 and part 3 is equal to part 2. This means that the overall equation for part 2 can be obtained by combining the ionic equations for parts 1 and 3. However, the equation you provided, NaOH + H + Cl = Na + Cl + H2O + 2 heats, does not accurately represent this.
It's important to note that in enthalpy labs, we are concerned with the heat involved in the reactions. In this case, both part 2 and part 3 involve the release of heat.