# Titrations and pka's

## 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

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.

Last edited:

symbolipoint
Homework Helper
Gold Member
K = $$\frac{(H)(A)}{HA}$$ , ignoring any charges. Monofunctional weak acid, HA.

-log(K) = -(log(H) + log(A/HA)

pK = pH + log(A/HA)

What happens at half-neutralization of the acid? [A]/[HA] = 1 (Do you understand this?) The concentration of unneutralized acid and neutralized acid is the same, equal so their ratio is 1.

pK = pH + log(1) = pH + 0 = pH.
pK=pH at half-neutralization.

Borek
Mentor
Given: titrant molarity is .2250 M ; 50 mL of analyte present ; 20 mL of titrant ; initial pH of 3.0

Is it all that is given? Does it mean 20 mL used to titrate 50 mL? If so, question reads: what is pKa of the acid if 0.09M solution has pH 3.0, and 5.0 seems to be OK.

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