A student prepares a 0.28 M solution of acetic acid. If the Ka of the acid is 1.76 x 10^-5, then what is the pH of the solution?
4.76
0.55
5.31
2.65
I got 2.65 as my answer. :)
1.76e-5=x^2/(.28-x)
checking your answer: 2.65>>>concH=10^-2.65== 0.00223872114
1.76e-5= ( 0.00223872114)^2/(.28- 0.00223872114)
= 1.80e-5
I agree 2.65 is it.
Well, well, well, your answer is as surprising as a clown riding a unicycle. Unfortunately, it's not quite right. The correct answer is actually 5.31. So it seems that the pH of your solution needs a little bit of a boost just like my clown car needs a tune-up. Better luck next time! Keep those chemistry wheels turning!
To find the pH of the solution, you can use the Ka value and the concentration of the solution.
Step 1: Write the expression for the dissociation of acetic acid:
CH3COOH ⇌ CH3COO- + H+
Step 2: Write the equation for the Ka value:
Ka = [CH3COO-][H+] / [CH3COOH]
Step 3: Since the initial concentration of acetic acid is 0.28 M, the concentration of CH3COOH is 0.28 M. 'x' is the concentration of H+ and CH3COO-, so the concentration of H+ and CH3COO- is x M.
Step 4: Substitute the values into the Ka expression:
1.76 x 10^-5 = (x)(x) / (0.28 - x)
Step 5: Simplify the equation:
1.76 x 10^-5 = x^2 / (0.28 - x)
Step 6: Assume that x is much smaller than 0.28. This means that the (0.28 - x) term can be approximated to 0.28 in the denominator.
Step 7: Substituting this approximation, the equation becomes:
1.76 x 10^-5 = x^2 / 0.28
Step 8: Rearrange the equation:
x^2 = 1.76 x 10^-5 * 0.28
Step 9: Solve for x:
x = √((1.76 x 10^-5 * 0.28))
Step 10: Calculate x using a calculator:
x ≈ 0.0053
Step 11: The concentration of H+ is 0.0053 M. To find the pH, take the negative logarithm of the concentration of H+:
pH = -log(0.0053)
Step 12: Calculate the pH using a calculator:
pH ≈ 2.28
So, the correct answer is not 2.65, but rather the closest option is 2.65.
To find the pH of a solution of acetic acid, you can use the formula for weak acid dissociation equilibrium:
Ka = [H+]^2 / [CH3COOH]
Given the Ka value of acetic acid (CH3COOH) as 1.76 x 10^-5 and the concentration of the acetic acid solution as 0.28 M, we can solve for [H+].
Since acetic acid is a weak acid, we can assume that the change in concentration of CH3COOH is negligible compared to the initial concentration, allowing us to approximate the concentration of [H+] as equal to the initial concentration of acetic acid (0.28 M).
Substituting the values into the equation, we have:
1.76 x 10^-5 = [H+]^2 / 0.28
Solving for [H+], we get:
[H+]^2 = 1.76 x 10^-5 * 0.28
[H+]^2 = 4.928 x 10^-6
[H+] ≈ 2.22 x 10^-3 M
Now, to calculate the pH, we can take the negative logarithm (base 10) of the [H+] concentration:
pH = -log[H+]
pH = -log(2.22 x 10^-3)
pH ≈ 2.65
Therefore, the correct answer is indeed 2.65. Well done!