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.
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)
4 answers
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.
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%.
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%.
2 answers