I did an experiment to study the effects of changes in conditions on an equilibrium system. The first system was the dissociation of acetic acid.
HC2H3O2 -> H^+^ + C2H3O2^-^
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We had to add 2mL of acetic acid into a test tube and we had to test the pH of the solution and record the color.
The solution was clear at first and then on the indicator paper the color was red with a pH of 4.
Then to this solution we added solid NaC2H3O2 and tested the pH again. This time the pH was 6 and the color was orange.
The question that I have to answer is did the equilibrium shift and in which direction?
The equilibrium did shift and I think that it shifted towards the reactant side. This would then be an endothermic process because products were removed??
Also, how did the pH of the solution change after adding NaC2H3O2 to it? Identify which ion is being added to the solution and how it affects the concentration of other ions present in the solution.
The pH increased. I am not sure about the ions. Na was added to the solution which increased the concentration.
Am I somewhat correct with these answers?? Could someone please explain? Thank you!!
A system of acetic acid will shift to the left if acetate ion is added to it; however, I don't think that tells you it is an endothermic reaction.
Well Na ions were added so would that still shift to the left? How do you know if it's endo or exothermic I have to include that in my answer?
Your understanding is partly correct, but let me clarify and explain the concepts to you in more detail.
When you added solid NaC2H3O2 to the acetic acid solution, you introduced the acetate ion (C2H3O2-) into the solution. The reaction that occurs is:
HC2H3O2 + NaC2H3O2 -> H^+ + Na^+ + C2H3O2^-
This is a dissociation reaction, where the sodium acetate salt (NaC2H3O2) breaks down into its ions in water. The acetate ion (C2H3O2-) acts as a base, accepting a proton from the hydronium ion (H+) to form acetic acid (HC2H3O2).
By adding the acetate ion, you increased the concentration of the acetate ions compared to the hydronium ions in the solution. The presence of excess acetate ions pushes the equilibrium in the direction that will consume those additional ions, which is towards the reactant side. In other words, the equilibrium shifts to the left to produce more acetic acid (HC2H3O2) and hydronium ions (H+).
Since acetic acid is a weak acid, it doesn't fully dissociate in water, resulting in the formation of some hydronium ions. By shifting the equilibrium to the left, you are essentially increasing the concentration of acetic acid and hydronium ions, leading to a decrease in pH. This explains the decrease in pH from 4 to 6 after adding NaC2H3O2 to the solution.
In summary, the equilibrium shifted towards the reactant side (to the left) after adding NaC2H3O2, causing an increase in the concentration of acetic acid and hydronium ions, and thereby decreasing the pH of the solution. The acetate ions act as a base, accepting protons and favoring the formation of acetic acid.
Yes, you are partially correct in your answers. Let's break it down step by step:
1. Did the equilibrium shift and in which direction?
In this case, you are studying the dissociation of acetic acid. When acetic acid dissociates, it forms hydrogen ions (H+) and acetate ions (C2H3O2-). The equilibrium expression for this reaction can be represented as:
HC2H3O2 ⇌ H+ + C2H3O2-
In the initial solution, the pH was measured to be 4, indicating a relatively high concentration of H+ ions. When you added solid NaC2H3O2 to the solution, it dissolved to form sodium ions (Na+) and acetate ions (C2H3O2-). The addition of NaC2H3O2 provides more acetate ions to the solution. According to Le Chatelier's principle, when you increase the concentration of one of the products (C2H3O2- in this case), the equilibrium will shift to the left to counteract the change. As a result, more acetic acid (HC2H3O2) will be formed, which means more hydrogen ions (H+). This shift towards the reactant side causes the pH to decrease.
So, in your experiment, since the initial pH was 4 and it increased to 6 after the addition of solid NaC2H3O2, the equilibrium shifted to the left, favoring the formation of more acetic acid and hydrogen ions (H+). Therefore, you are correct that the equilibrium shifted towards the reactant side.
2. How did the pH of the solution change after adding NaC2H3O2, and which ion was added and how it affected the concentration of other ions?
You correctly mentioned that the pH increased after adding NaC2H3O2. This is because, as mentioned earlier, the addition of NaC2H3O2 provides more acetate ions (C2H3O2-) to the solution. Acetate ions are the conjugate base of acetic acid and are capable of accepting hydrogen ions (H+) in solution. By increasing the concentration of acetate ions, the solution's ability to accept hydrogen ions is enhanced, resulting in a higher pH.
Regarding the ions, when NaC2H3O2 dissolves in water, it dissociates into sodium ions (Na+) and acetate ions (C2H3O2-). The addition of Na+ does not directly affect the concentration of H+ ions. However, as mentioned earlier, the increased concentration of acetate ions (C2H3O2-) influences the equilibrium to shift to the left, resulting in an increase in the concentration of H+ ions, which lowers the pH.
In summary, you are correct that the pH increased after adding NaC2H3O2 and that Na+ was added to the solution. However, the shift towards the reactant side favored the formation of more acetic acid and hydrogen ions, which led to a decrease in pH, not an increase. Additionally, the increased concentration of acetate ions affected the equilibrium and indirectly influenced the concentration of H+ ions and the pH of the solution.