Net ionic equations (NaOH; HCl; H2O)

In summary, the conversation discussed the completion of a Grade 12 experiment and the confusion in calculating the net ionic equation and ΔH for three reactions involving NaOH and HCl. The instructions were also provided, stating that the solutions can be assumed to have a density of 1.00 g/mL. The conversation also included the writer's attempt at writing the net ionic equations for each reaction and a request for help. The expert advised to start by writing dissociated substances as ions and cancelling ions when they appear in the same quantity on both sides of the balanced equation. The expert also noted that the given information was incomplete and confusing, as it did not specify if the experiment was to calculate expected temperature change or to use measured data to
  • #1
berry2
5
0
Hi,

I completed a Grade 12 experiment and now I am doing some calculations about it but I am confused.

Here is the question: "Write the net ionic equation for each reaction, and note the value of ΔH for each reaction. e.g. H+(aq) + OH-(aq) --> H2O(l)
ΔH = -45 kJ/mol HaOH

I had three reactions.
1. Heat of dissolution of NaOH; 200 mL of labelled water + 5.5 g of crystal NaOH

2. Heat of reaction between aqueous NaOH and aqueous HCl; 100 mL of 1.0 mol/L HCl + 100 mL of 1.0 mol/L NaOH

3. Heat of reaction between solid NaOH and aqueous HCl; 200 mL of 1.0 mol/L HCl + 5.5g of crystal NaOH


These were the instructions also given: "Since all the solutions are dilute, then the density of each solution can be assumed to be 1.00 g/mL. So, 100 mL has a mass of 100 g."

So for the first reaction, would the net ionic equation be something like this:
1. 200 g H2O(l) + 5.5 g NaOH(s) --> I don't know what the product would be?
2. 100 g HCl(aq) + 100 g NaOH(aq) --> NaCl + H2O?
3. 200 g of HCl(aq) + 5.5 g NaOH(s) --> NaCl + H2O?

Any help would really be appreciated!:confused:
Thanks
 
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  • #2
Whenever you are told to write the net ionic reaction you should start by writing every dissociated substance as ions - so not NaOH(aq), but Na+(aq) + OH-(aq) and so on. Then cancel ions whenever they appear in the same quantity on both sides of the balanced equation.

In the first reaction you have NaOH(s) on the left, so the reaction is just that of a dissociation. Two other reaction equations are OK.

Other than that it is difficult to help, as you didn't give us complete information, and the information you gave is a bit confusing: you wrote you had "reactions", but apparently the experiment was about enthalpy changes, so you are either expected to calculate expected temperature change, or you measured the temperature change and you are expected to use the data to calculate ΔH. Note that these are actually limiting reagent problems - masses given suggest you are expected to calculate amount of substances that reacted.
 

Related to Net ionic equations (NaOH; HCl; H2O)

What is a net ionic equation in chemistry?

A net ionic equation is a chemical equation that represents only the species that actively participate in a chemical reaction, excluding spectator ions that do not undergo any change during the reaction. It provides a simplified view of the chemical changes occurring in a reaction.

How do you write a net ionic equation?

To write a net ionic equation, follow these steps:

  1. Write the balanced chemical equation for the reaction, including all reactants and products.
  2. Identify and write the complete ionic equation, showing all ions present in their dissociated form.
  3. Identify and cancel out spectator ions, which are ions that appear on both sides of the equation without undergoing a chemical change.
  4. Write the net ionic equation, which includes only the species that actively participate in the reaction.

What happens when NaOH reacts with HCl in water (H2O)?

When NaOH (sodium hydroxide) reacts with HCl (hydrochloric acid) in water (H2O), a neutralization reaction occurs. Sodium hydroxide (NaOH) is a strong base, and hydrochloric acid (HCl) is a strong acid. They react to form water and a salt, sodium chloride (NaCl).

Can you provide the balanced chemical equation for the reaction between NaOH and HCl in water?

Certainly! The balanced chemical equation for the reaction is: NaOH + HCl → NaCl + H2O

What is the complete ionic equation for this reaction?

The complete ionic equation for the reaction between NaOH and HCl in water is: Na⁺ + OH⁻ + H⁺ + Cl⁻ → Na⁺ + Cl⁻ + H₂O

What are the spectator ions in this reaction, and how do you identify them?

In this reaction, the spectator ions are Na⁺ and Cl⁻. You can identify spectator ions by comparing the ions present on both sides of the equation. If an ion appears unchanged on both sides, it is a spectator ion and can be canceled out in the net ionic equation.

What is the net ionic equation for the reaction between NaOH and HCl in water?

The net ionic equation for the reaction, after canceling out the spectator ions, is: OH⁻ + H⁺ → H₂O

What does the net ionic equation reveal about the reaction?

The net ionic equation reveals that in the reaction between NaOH and HCl in water, hydroxide ions (OH⁻) from the base react with hydrogen ions (H⁺) from the acid to form water (H₂O). This represents the essential chemical change that occurs in the neutralization reaction.

Why is it useful to write net ionic equations in chemical reactions?

Writing net ionic equations is useful because they provide a clearer and more concise representation of the chemical changes occurring in a reaction. They focus on the species directly involved in the reaction, making it easier to understand and analyze the essential chemistry taking place while excluding irrelevant ions (spectators).

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