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## Homework Statement

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

## Homework Equations

M1*V1=M2*V2

## 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.

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