What is the pH of a 0.25 M acetic acid solution if the Ka of acetic acid is 1.8 x 10-5?
2.67
0.60
5.34
11.33
I got A, 2.67. I think this is correct as well since acetic acid is a weaker acid and the ph # would make sense.
CH3CO2H → CH3CO2− + H+
[CH3CO2−] [H+] / [CH3CO2H] = 1.8E-5
x^2 / (.25 - x) = 1.8E-5 ... x^2 + 1.8E-5 x - 4.5E-6 = 0 ... x = 2.11E-3
log(2.11E-3) = 2.67 ... good choice
Scott you a Dawg brother
Well, well, well, looks like we've got a chemistry whiz here! You're absolutely correct! Acetic acid, being a weak acid, will not fully ionize in solution. The pH of a 0.25 M acetic acid solution can be calculated using the Ka value. So, kudos to you for picking A, 2.67! You've kicked some acid with that answer! Keep up the great work!
To find the pH of a solution of acetic acid, we need to consider its dissociation constant (Ka).
The dissociation of acetic acid (CH3COOH) in water can be represented as follows:
CH3COOH + H2O ⇌ CH3COO- + H3O+
The Ka expression for this reaction is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
The problem states that the concentration of acetic acid is 0.25 M and the Ka is 1.8 x 10^(-5). We can assume that the concentration of [CH3COO-] and [H3O+] are equal, as acetic acid is a weak acid.
Let x be the concentration of [CH3COO-] and [H3O+]. Since the initial concentration of acetic acid is 0.25 M and it is a weak acid, we can assume that the change in concentration of [CH3COO-] and [H3O+] is very small compared to the initial concentration. Therefore, we can approximate the concentration of [CH3COO-] and [H3O+] as x.
Using the Ka expression, we have:
1.8 x 10^(-5) = (x)(x) / 0.25
Simplifying the expression:
1.8 x 10^(-5) = x^2 / 0.25
Rearranging the equation:
x^2 = (1.8 x 10^(-5))(0.25)
x^2 = 4.5 x 10^(-6)
Taking the square root of both sides to solve for x:
x ≈ 2.12 x 10^(-3)
Now we can find the pH of the solution using the equation for the pH of a solution:
pH = -log[H3O+]
Since [H3O+] ≈ x = 2.12 x 10^(-3), we can substitute it into the equation to find the pH:
pH ≈ -log(2.12 x 10^(-3))
pH ≈ 2.67
So, you are correct! The pH of the 0.25 M acetic acid solution is approximately 2.67.
To determine the pH of a solution, we can rely on the equation for the dissociation of acetic acid:
CH3COOH ⇌ CH3COO- + H+
First, we need to calculate the concentration of H+ ions produced from the dissociation of acetic acid. This can be done by taking the square root of the product of the initial concentration of the acid (0.25 M) and the acid dissociation constant (Ka = 1.8 x 10^-5):
[H+] = √(0.25 M * 1.8 x 10^-5)
[H+] = 0.006 M
Now that we have the concentration of H+ ions, we can calculate the pH by taking the negative logarithm of the H+ concentration:
pH = -log([H+])
pH = -log(0.006)
pH ≈ 2.62
Based on the given choices, it seems that the correct answer is not A (2.67), but rather a value very close to 2.62. Therefore, the most appropriate option is possibly "5.34".