1. The problem statement, all variables and given/known data 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 2. Relevant equations M1*V1=M2*V2 3. The attempt at a solution M1*V1=M2*V2 (.2250M)*(20mL)=M2*(50mL) 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.