Given acidity constant,Ka of benzoic acid is 6.28 X 10^-5. pH of 0.15 molar solution of this acid is ?
Here is my method:
Benzoic acid is a weak acid,hence it dissociates very little. So:
C6H5COOH---> C6H5COO- + H+
[H+] and [C6H5COO-] are yet to be calculated,so let them be x
[C6H5COOH] = 0.15-x
Since Ka= 6.28 X 10^-5 mol/L,using the Kc equation:
Ka = x.x/(0.15-x)
Assuming x<< 0.15,simplifying the eqn becomes : Ka = x.x/(0.15)
Solving the equation,x value is 3.069201851 X 10^-3 mol/L = [H+].
pH = -log[H+] = 2.51. Is this correct?
Thank you
I agree. Thanks for showing your work. That helps a lot.
Well, your calculations seem to be in the right direction, but let's add a little humor to it:
So, you have your benzoic acid doing a little dissociation dance, like "C6H5COOH---> C6H5COO- + H+". It's a weak acid, so it doesn't really feel like breaking up much. It's like those couples who can't decide if they're really done with each other, you know?
Now, you've got your concentration of benzoic acid as 0.15 M, but since it's not fully dissociated, you subtract x from that to get the concentration of your undissociated acid, which is 0.15 - x. You've got the idea!
Now, you use the acidity constant Ka, which is like a popularity measure for the acid. It tells you how likely it is to lose that H+. It's a small Ka, so it's not super popular, but hey, it's not the loneliest acid in the party either.
Using the equation Ka = x^2 / (0.15), you find the value of x to be 3.069201851 X 10^-3 M. That's the concentration of our little H+ party crasher.
Finally, you calculate pH, also known as "the power of H+". You take the negative logarithm of the H+ concentration, and *ding ding ding*, you get a pH value of 2.51. Ta-da!
So, yes, your calculation seems correct! Great job, scientist! I hope my humor added a little fun to the process. Keep up the good work!
Your method is correct, and the calculation is accurate. The pH of a 0.15 molar solution of benzoic acid is indeed 2.51. Well done!
Yes, your method is correct. Let's go through the steps to calculate the pH of the 0.15 molar solution of benzoic acid.
1. Write down the dissociation equation for benzoic acid: C6H5COOH → C6H5COO- + H+
2. Assume that x represents the concentration of both [H+] and [C6H5COO-].
3. Since benzoic acid is a weak acid, it dissociates very little, so the initial concentration of benzoic acid is equal to the concentration of undissociated acid: [C6H5COOH] = 0.15 - x.
4. The equilibrium constant expression for the dissociation of benzoic acid is Ka = [C6H5COO-][H+]/[C6H5COOH].
5. Substituting the values, we get Ka = x*x/(0.15-x).
6. Since the value of x is expected to be much smaller compared to 0.15, we can simplify the equation to Ka = x*x/0.15.
7. Rearranging the equation, we find that x^2 = Ka * 0.15.
8. Solving for x, we get x ≈ 3.069201851 × 10^-3 mol/L, which represents the concentration of [H+].
9. Finally, we can calculate the pH using the formula pH = -log[H+]. Plugging in the value of [H+], we find that the pH is approximately 2.51.
Therefore, based on your calculations, the pH of the 0.15 molar solution of benzoic acid is indeed 2.51.