Asked by Summer

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?
Answered by DrBob222
(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 Henderson-Hasselbalch 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.
Answered by Summer
What would the Ka for this be?
Answered by Summer
Oops, sorry. I understand now, thank you.
Answered by DrBob222
I calculated pKa = 4.25 so Ka = 5.624E-5
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