Weak acid analyte with strong base titrant.
What is the pKa of the analyte in this titration to the nearest 0.5?
Given: titrant molarity is .2250 M ; 50 mL of analyte present ; 20 mL of titrant ; initial pH of 3.0
The Attempt at a Solution
M2 = .09M
Knowing the molarity, I made a table and got:
Ka = x^2/(.09-x), where x is the concentration of H^+ ions.
-log(x) = 3.0
x = .001
Ka = (.001)^2/.09 = 1.11e-5
pKa = -log(1.11e-5) = 4.95 ----------> Since to nearest 0.5, pKa = 5.0
This last part is where I'm confused. I submitted this answer for my homework and it said it was incorrect. Is this not what they meant by nearest 0.5? Any help is much appreciated.