What would be the ph of a 0.20M CH3COOH solution if no salt were present?
Vinegar is acetic acid, which I will write as CH3COOH. It's the right end H that reacts with the NaOH.
CH3COOH + NaOH ==> CH3COONa + HOH.
SO, mols NaOH = L x M = 0.0267 L x 0.600 M = ?? mols NaOH.
The whole idea of titration is that the mols of the titrant will tell us the amount of the unknown material in the sample.
So we have ?? mols NaOH.
You can see we replaced the 1 H on CH3COOH with OH of the NaOH, so we have 1 mol CH3COOH per 1 mol NaOH. That means we must ALSO have ?? mols CH3COOH.
Since mols = M x L, you know mols CH3COOH, you know L CH3COOH from the problem, solve for M, the molarity.
It can't get much simpler than that; you may be making it too hard. # mols = M x L is all you need to know. Maybe Idk
2.27
Well, well, looks like CH3COOH is up to some pH mischief, huh? Alright, let's tackle this question with a dash of humor! If no salt were present in a 0.20M CH3COOH solution, it would be acidic enough to make a lemon pucker up. The pH value would be around 2.4 - acidic, but not quite as intense as trying to eat a warhead candy! Now, I hope that acidity level didn't give you heartburn!
To determine the pH of a 0.20M CH3COOH (acetic acid) solution without any salt present, we need to calculate the acid dissociation constant (Ka) of acetic acid and its concentration in solution.
The Ka value for acetic acid (CH3COOH) is approximately 1.8 x 10^-5 at 25°C.
The dissociation of acetic acid (CH3COOH) in water can be represented by the equation:
CH3COOH ⇌ CH3COO- + H+
CH3COOH is a weak acid, meaning it only partially dissociates in water. To find the concentration of the hydrogen ion (H+), we have to assume that the acetic acid completely dissociates.
So, at equilibrium, the concentration of H+ will be equal to the concentration of CH3COOH that dissociates. Given that the initial concentration of CH3COOH is 0.20M, we can assume that [H+] will also be 0.20M.
Using the formula for pH:
pH = -log [H+]
pH = -log (0.20) ≈ 0.70
Therefore, the pH of a 0.20M CH3COOH solution without any salt present is approximately 0.70.
To determine the pH of a 0.20M CH3COOH (acetic acid) solution, we need to consider its ionization in water. Acetic acid is a weak acid and partially ionizes in water, producing both hydronium ions (H3O+) and acetate ions (CH3COO-).
The dissociation equation for acetic acid can be written as follows:
CH3COOH ⇌ CH3COO- + H3O+
Since acetic acid is a weak acid, it does not completely dissociate in water. Instead, the extent of ionization is determined by the acid dissociation constant, known as the Ka value.
In this case, the Ka value for acetic acid is 1.8 x 10-5 (at 25°C).
To calculate the pH of the acetic acid solution, we need to find the concentration of H3O+ ions. This can be achieved by determining the concentration of the acetate ion using the equation:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Since initially, only acetic acid is present, and there is no salt, the concentrations of both acetate ion and hydronium ion will be equal. Hence, we can consider the concentration of H3O+ as 'x'.
Using the Ka expression and given values, we can set up the equation:
1.8 x 10-5 = (x)(x) / (0.20 - x)
Solve this equation using quadratic formula or approximation techniques to find the value of 'x', which represents the concentration of H3O+ ions.
Once you determine 'x', you can calculate the pH using the equation:
pH = -log[H3O+]
The resulting pH value will represent the acidity of the acetic acid solution.