Consider the titration of 100.0 mL of 0.260 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.130 M NaOH. Calculate the pH of the resulting solution after each of the following volumes of NaOH has been added. (Assume that all solutions are at 25°C.)

Question

(a) 0.0 mL

(b) 50.0 mL

(c) 100.0 mL

(d) 150.0 mL

(e) 200.0 mL

(f) 250.0 mL

I have done a through d already and got them right,
However when I get to e I start having problems, theres something I am missing.

Here is what I have tried:
Initial Concentration of Acid= (.100*.260)=.026 M
Initial Concentration of Base=
(.200*.130)=.026 M

Then:
Kb= 1.0e-14/1.3e-5= 7.69e-10
x^2/.026= 7.69e-10

x=sqrt(.026*7.69e-10)=4.47e-6
pOH= -log(4.47e-6)=5.35
pH=14-5.35=8.65

The homework system says I am incorrect so I would like some assistance in figuring out what I am doing wrong.

Thank You

Answer 1

All of that looks good EXCEPT for using 0.026M for concentration of the propanoate ion. That is not 0.026 because it has been diluted by the addition of the NaOH.
You started with 100 mL of 0.026M acid, when you've added 200 mL of NaOH the salt concn is now 0.026 x (100/300) = ?
Thanks for showing your work. It makes it easy to spot the problem and I don't need to work all of that stuff you already know how to do.