I am trying to solve a buffer problem. I know I need to use the Henderson-Hasselblach equation but I am not getting the correct answer.
The problem is: Propionic acid is a weak monoprotic acid with Ka = 1.3×10-5 M. NaOH(s) was gradually added to 1.00 L of 1.29×10-2 M propionic acid. Calculate the pH of the solution after the addition of 3.87×10-3 mol of NaOH(s).
I have tried -log(1.3E-5)+ log(3.87E-3/1.2E-2)
You don't need it. Just calculate the new molarity of the acid after addition of NaOH, and then determine the molarity of the acid in solution.
How many moles of propionic acid are left after addition, of the strong base?
In this case, Molarity=moles
1.29 x 10^-2 M=1.29 x 10^-2 moles
1.29 x 10^-2 moles-3.87 x-3.87 x 10^-3 moles=9.03 x 10^-3 moles of propionic acid.
9.03 x 10^-3 moles of propionic acid=9.03 x 10^-3 M of propionic acid
HA -----> H^+ + A^-
Ka=1.3×10-5=[H^+][A^-]/[HA]
or
Ka=[x][x]/[9.03 x 10^-3 M]
let
x=H^+ and A^-
and
HA=9.03 x 10^-3 M
Solve for x:
x=Sqrt*[Ka*9.03 x 10^-3 M]
Check 5% dissociation after you are done, and if it is greater than 5%, you will have to use the quadratic equation to solve for x.
pH=-log[H^+]
you know that x=[H^+], so solve for pH.
Thank you so much!
Shouldn't you have used the salt for the concentration of (A^-) and not x which is the same as (H^+).
From your work,
Ka=[x][x]/[9.03 x 10^-3 M]
would be
Ka = (x)(x+salt)/(HA) where salt is 3.87E-3 M.
I do apologize about that. I totally forgot about the salt.
Ignore the x for the salt in the setup that Dr. bob222 gave you because we are assuming that dissociation is small and solve for x. When I gave you the setup, I ignored the x in the denominator assuming dissociation is small, which is needed for the numerator as well. Without taking short cuts, the setup is as followed:
Ka=1.3×10-5=[H^+][A^-]/[HA]
Ka=[x][x+3.87E-3 M ]/[9.03 x 10^-3 M-x]
Assuming dissociation is small, the equation becomes:
Ka=[x][3.87E-3 M ]/[9.03 x 10^-3 M]
Solving for x:
x=Sqrt*{[Ka*9.03 x 10^-3 M]/3.87E-3 M}
I apologize about that. I took several shortcuts and I did not do an ICE chart, which would have prevented me from making that mistake.
Still check to see if dissociation is greater than 5%.
To solve this buffer problem using the Henderson-Hasselbalch equation, you're on the right track, but there are a couple of things to correct in your calculation.
First, let's define the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Here:
pH = unknown
pKa = -log(Ka) = -log(1.3×10^-5)
[A-] = concentration of the conjugate base (in this case, the concentration of the propionate ion, C3H5O2-)
[HA] = concentration of the acid (in this case, the concentration of propionic acid, C3H6O2)
To calculate the concentrations [A-] and [HA]:
[A-] = initial concentration of the acid + moles of the acid transformed into the conjugate base (propionate)
[HA] = initial concentration of the acid - moles of the acid transformed into the conjugate base (propionate)
Given:
Initial concentration of propionic acid (C3H6O2) = 1.29×10^-2 M
Moles of NaOH(s) added = 3.87×10^-3 mol
Using these values, we can calculate [A-] and [HA]:
[A-] = (1.29×10^-2 M) + (3.87×10^-3 mol / 1.00 L)
[HA] = (1.29×10^-2 M) - (3.87×10^-3 mol / 1.00 L)
Now, substitute these values into the Henderson-Hasselbalch equation to calculate pH:
pH = -log(1.3×10^-5) + log([A-] / [HA])
Plug in the values for [A-] and [HA] into the equation and calculate:
pH = -log(1.3×10^-5) + log(([A-]) / ([HA]))
Now, evaluate the expression inside the logarithm:
[A-] = (1.29×10^-2) + (3.87×10^-3 / 1.00) = (1.29×10^-2) + 3.87×10^-3 = (1.29×10^-2) + (3.87×10^-3) = (1.29×10^-2) + (3.87×10^-3) = (1.29×10^-2) + (3.87×10^-3) = 1.29×10^-2 + 3.87×10^-3 = 16.29×10^-3 = 1.629×10^-2 M
[HA] = (1.29×10^-2) - (3.87×10^-3 / 1.00) = (1.29×10^-2) - 3.87×10^-3 = 1.29×10^-2 - 3.87×10^-3 = 1.29×10^-2 - 3.87×10^-3 = 12.9×10^-3 = 1.29×10^-2 M
Now, substitute these values into the Henderson-Hasselbalch equation:
pH = -log(1.3×10^-5) + log(1.629×10^-2 / 1.29×10^-2)
Evaluate the logarithm:
pH = -log(1.3×10^-5) + log(1.629×10^-2 / 1.29×10^-2) = -log(1.3×10^-5) + log(1.26)
Finally, calculate the pH using a calculator:
pH ≈ -(-4.89) + 0.1 ≈ 4.89 + 0.1 ≈ 4.99
Therefore, the pH of the solution after the addition of 3.87×10^-3 mol of NaOH(s) is approximately 4.99.