posted by phil on .
What is the pH of a solution prepared by adding 4.000 grams of NaOH to a 50.00 mL of a buffer that is 2.00 molar both in acetic acid and sodium acetate? (Assume final volume is still 50.00 mL)
Use the Henderson-Hasselbalch equation. The addition of NaOH will decrease the acid by that many moles and increase the acetate by that many moles. Substitute into the HH equation and solve for pH. Post your work if you get stuck.
i did .1 mols of AC- + .1 mols of OH gives me .2 mols of AC.
the .2 mols of AC reacts with water
AC- + H20---> OH- + HAC
so the M is .2/.05 L = 4M
to solve for the OH conc i did X^2 / 4 = 5.6 x 10^-10. and solving for the pH i get 9.675
is this right?
You are exactly right. In fact, when I first responded to the problem I was not aware that the amount of NaOH added was enough that we no longer have a buffer but just sodium acetate salt, which of course, hydrolyzes. When I sat down to work the problem to explain what "you did wrong" I discovered what was going on. Your answer is right on the money.