Chemistry
posted by Summer on .
A solution that contains 2.047 x 10^1 M of acid, HA, and 1.808 x 10^1 M of its conjugate base, A, has a pH of 4.196. What is the pH after 1.629 x 10^3 mol NaOH is added to 5.961 x 10^1 L of this solution?

(HA) = 0.2047
(A^) = 0.1808
moles HA = 0.2047M*0.5961L = 0.12202
moles A^ = 0.1808 x 0.5961 = 0.10777
pH = pKa + log (base)/(acid)
The problem states the pH as well as the beginning HA and A. Substitute those and solve for pKa. Then look at the ICE chart below.
............HA + OH^ ==> A^ + H2O
initial.0.12202...0...0.10777....
add............0.001629............
change...001629..001629..+.001629.....
equil......?......0.......?
Fill in the ? spots and substitute into the HendersonHasselbalch equation to solve for new pH. Use pKa from the first calculation above. Post your work if you need additional help. I have carried more digits than allowed; you should round as needed. 
What would the Ka for this be?

Oops, sorry. I understand now, thank you.

I calculated pKa = 4.25 so Ka = 5.624E5