Calculate the pH of a [Acid] M aqueous solution of each of the hydrochloric and acetic acids:
Acid [Acid] pH
HCl 0.10
HC2H3O2 0.10
I know that HCl is just -log(0.10) which is 1, but can't get the pH of HC2H3O2.
You need the molarity.
pH = -log(H^+)
1, 2,87
i'm doing this problem too =)
Molarity for both hydrochloric and acetic acids are 0.10. For hydrochloric acid, i used the formula: pH = -log (0.10) = 1.
But should we do something else for acetic acid? because its pH is not 1.
never mind, I found out how to do this =)
(the post above was me by the way).
set up I.C.E. table for the equilibrium expression. Since we know that Ka value for acetic acid is about 1.76 x 10^-5, we can solve for x (which is [H+] in this case).
Ka = x^2 / 0.10 - x => approximation: 1.76 x 10^-5 = X^2 / 0.10. Solve for x. Then use pH = -log [H+] to find pH of acetic acid =).
To calculate the pH of an aqueous solution of an acid, you need to know the concentration of the acid and the dissociation constant of the acid. In this case, you know the concentration of both hydrochloric acid (HCl) and acetic acid (HC2H3O2) is 0.10 M.
To get the pH of hydrochloric acid (HCl), you can directly calculate it using the formula -log[H+]. Since hydrochloric acid is a strong acid, it dissociates almost completely in water to form hydronium ions (H+). Therefore, the concentration of H+ ions is equal to the concentration of HCl. Hence, the pH = -log(0.10) = 1.
On the other hand, acetic acid (HC2H3O2) is a weak acid. It only partially dissociates in water, meaning it does not fully produce H+ ions. To calculate the pH of a weak acid solution, we need to consider the equilibrium constant expression for the dissociation of the acid.
For acetic acid, its dissociation reaction is represented as follows:
HC2H3O2 ⇌ H+ + C2H3O2-
The equilibrium constant for this reaction is represented as Ka. The dissociation constant (Ka) of acetic acid is approximately 1.8 x 10^-5 M.
To calculate the concentration of H+ ions in the solution, we can assume that the dissociation of acetic acid is x (where x is small) and since it is a 0.10 M solution, the concentration of the undissociated form will be (0.10 - x).
Using the equilibrium constant expression for the weak acid, we can write:
Ka = [H+][C2H3O2-] / [HC2H3O2]
Since HC2H3O2 is a monoprotic acid, the concentration of [C2H3O2-] is also equal to the concentration of H+ ions. So we can substitute [C2H3O2-] as x in the expression:
Ka = x * x / (0.10 - x)
Since the value of x (concentration of H+ ions) is small compared to 0.10, the approximation can be made:
Ka ≈ x * x / 0.10
Rearranging the equation and solving for x:
x * x = Ka * 0.10
x = √(Ka * 0.10)
Using the given Ka value of 1.8 x 10^-5 M:
x = √(1.8 x 10^-5 * 0.10) ≈ 0.004
Now, to get the pH, we take -log(x):
pH ≈ -log(0.004) ≈ 2.40
Therefore, the pH of a 0.10 M acetic acid (HC2H3O2) solution is approximately 2.40.