Vinegar is a 5.0% solution by weight of acetic acid in water. Given that the pH for vinegar is 2.41 and that Ka= 1.8E-5 and the density s 1.00g/mL, what is the percent dissociation of acetic acid in vinegar?
I know the % dissociation equals the amount which did dissociate 5.0%divided by the amount which could have dissociated in this case 1.00 M because it says in the question that this is a 1.00 M solution of CH3COOH. Am I on the right track?
Let's take a humorous approach to this question!
Well, you're on track, but let me entertain you with a circus analogy to make things clearer. Imagine you have a crowd of 100 people sitting in the stands, representing the total amount of acetic acid in vinegar. Now, the percentage dissociation is like the number of people who stand up and leave the stands to join the circus act.
In this case, the "crowd" is 1.00 M acetic acid, and only a certain percentage dissociates. We want to find that percentage.
We know that vinegar has a concentration of 5.0% acetic acid, which means our crowd size is 5 people. However, not all of them will be bold enough to leave the stands and join the circus. The pH of 2.41 tells us that only a small fraction of the crowd members are brave enough to leave.
Now, here comes the trick! The Ka value of 1.8E-5 represents the bravery index of each person. The higher the value, the braver they are, and the more likely they are to dissociate.
To find the percentage dissociation, divide the dissociated amount (the brave circus performers) by the total amount present (the crowd in the stands), and multiply by 100 to get a percentage.
So, using the given values, you can calculate the percentage dissociation of acetic acid in vinegar. Happy calculations, and may your chemistry circus show be a success!
You are partially on the right track. To calculate the percent dissociation of acetic acid in vinegar, you can use the expression:
% dissociation = (concentration of dissociated acid / initial concentration of acid) x 100
In this case, the initial concentration of acetic acid is not 1.00 M, but rather 5.0% by weight, which means 5.0 g of acetic acid in 100 mL (or 100 g) of vinegar. Since the density of vinegar is 1.00 g/mL, we have 5.0 g of acetic acid in 100 mL.
First, let's calculate the molarity (M) of acetic acid in vinegar:
Assuming the molar mass of acetic acid (CH3COOH) is 60.05 g/mol, we can calculate the number of moles of acetic acid in 5.0 g:
moles of acetic acid = (mass of acetic acid / molar mass of acetic acid)
= (5.0 g / 60.05 g/mol)
≈ 0.083 moles
Next, let's convert mL to liters:
volume of vinegar = 100 mL
= 100/1000 L
= 0.1 L
Now, we can calculate the molarity (M) of acetic acid:
Molarity of acetic acid = (moles of acetic acid / volume of vinegar in liters)
= (0.083 moles / 0.1 L)
= 0.83 M
Since the equation for the dissociation of acetic acid is CH3COOH ⇌ CH3COO- + H+, and the acid is monoprotic, the concentration of the dissociated acid is equal to the concentration of H+ ions.
Using the pH of vinegar, we can calculate the concentration of H+ ions:
pH = -log[H+]
2.41 = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-2.41)
= 4.47 x 10^(-3) M
Now, we can calculate the percent dissociation:
% dissociation = ([H+] / initial concentration of acid) x 100
= (4.47 x 10^(-3) M / 0.83 M) x 100
≈ 0.54%
Therefore, the percent dissociation of acetic acid in vinegar is approximately 0.54%.
You're on the right track! To determine the percent dissociation of acetic acid in vinegar, you need to compare the concentration of dissociated acetic acid to the initial concentration of acetic acid.
In this case, we know that vinegar is a 5.0% solution by weight of acetic acid in water. However, we don't have the exact molarity of the acetic acid. So, let's calculate it:
1. Convert the percent by weight to grams per 100 mL (or g/mL) since the density is given as 1.00 g/mL.
5.0% of 100 mL = 5.0 g
2. Use the molar mass of acetic acid (CH3COOH = 60.05 g/mol) to convert grams to moles:
5.0 g CH3COOH * (1 mol CH3COOH / 60.05 g CH3COOH) = 0.0832 mol CH3COOH
3. Now, we can calculate the initial concentration (in M) by dividing moles by liters:
0.0832 mol CH3COOH / 0.1 L = 0.832 M CH3COOH
So, you were correct that this is a 0.832 M solution of acetic acid.
Next, we need to find the amount of acetic acid that dissociates by using the given pH and Ka value.
The pH of 2.41 allows us to calculate the hydronium ion concentration ([H3O+]). Since the solution is acidic and acetic acid is a weak acid, we can assume that [H3O+] is approximately equal to the concentration of CH3COOH that dissociates.
Using the equation for pH, we have:
pH = -log[H3O+]
2.41 = -log[H3O+]
[H3O+] = 10^(-pH)
[H3O+] = 10^(-2.41) = 4.26 x 10^(-3) M
Since [H3O+] is equivalent to [CH3COOH] dissociated, we can use this value to calculate the percent dissociation.
Percent dissociation = ([CH3COOH] dissociated / [CH3COOH] initial) x 100
Percent dissociation = (4.26 x 10^(-3) M / 0.832 M) x 100 = 0.51%
Therefore, the percent dissociation of acetic acid in vinegar is approximately 0.51%.
Yes and no.
Percent dissociation has little to do with 5%. That's 5% by mass CH3COOH in water BUT that doesn't tell you that it is or is not dissociated by any amount.
Let's call vinegar HAc to avoid writing CH3COOH for it. H stands for H (of the COOH part) and Ac stands for the remainder of the moelcule (CH3COO).
HAc ==> H^+ + Ac^-
%diss = [(H^+)/(HAc)]*100
pH = 2.41. You can calculate (H^+) from that. You have (HAc) (which you calculate from the 5% and the density but I don't know that it is 1M). Is there something else in the problem?