The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is
__________. The Ka of acetic acid is 1.76 × 10-5.
I hope the method above agrees with the answer you have found.
sorry, answer found. no longer need help..
ph=4.1
To find the pH of the solution, we need to use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is as follows:
pH = pKa + log([A-]/[HA])
In this case, the acid is acetic acid (CH3COOH), and its conjugate base is acetate ion (CH3COO-).
First, let's find the pKa of acetic acid.
pKa = -log(Ka)
pKa = -log(1.76 × 10-5)
pKa = 4.753
Next, let's calculate the ratio of [A-] (sodium acetate) to [HA] (acetic acid).
[A-]/[HA] = ([sodium acetate]) / ([acetic acid])
[A-]/[HA] = (0.172 M) / (0.818 M)
[A-]/[HA] ≈ 0.210
Now, we can substitute these values into the Henderson-Hasselbalch equation:
pH = 4.753 + log(0.210)
Using a calculator, we find that log(0.210) ≈ -0.677.
pH ≈ 4.753 - 0.677
pH ≈ 4.076
Therefore, the pH of the solution is approximately 4.076.
You are missing units for your Ka.
I am sure that I have answered this one recently. Anyway start from the equation
HAc <-> H+ + Ac-
so
Ka = [H+][Ac-]/[HAc]
at the start we have 0.181 M HAc
and 0.172 M NaAc
at equilibrium we have
(0.181-x)M HAc (some has dissociated)
(0.172+x)M Ac- (we now have more Ac- due to the dissociation)
x M H+
so we can write Ka
(0.172-x)(x)/(0.181-x)= 1.76 × 10-5
you can either solve the quadratic or we can say that x is small wrt 0.172 and 0.181, hence
(0.172)(x)/(0.181)= 1.76 × 10-5
find x
hence find the pH=-log(x)