titrations and pka's

[w3390]

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.

[symbolipoint]

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]

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.

--
ChemBuddy chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), stoichiometry calculator (http://www.chembuddy.com/?left=EBAS&right=equation-balancing-stoichiometry)
www.ph-meter.info (http://www.ph-meter.info) - ph meter (http://www.ph-meter.info), ph electrode (http://www.ph-meter.info/pH-electrode)