Hi, i was wondering if you can help with my titration lab i have due today. a) p

Question

Hi, i was wondering if you can help with my titration lab i have due today.
a) part 1 i need to find the original molarity of the acidic acid solution.
b) find experimental value of Ka (and pKa) of acidic acid based in initial pH
c) initial value of Ka (and pKa) based on the pH at the half equivalence point
d) % error for all pKa values.
I have an initial pH of 2.88 (that i measured in my lab with a pH meter). i used 50.0mL of acetic acid with a molarity of 0.1M. I have a base NaOH of 0.219M. The endpoint was at 31.30mL with a pH of 9.90. If anything is missing let me know please. Thanks

Explanation / Answer

A) Original molarity of acetic acid in solution

moles NaOH used = 0.219 M x 31.30 ml = 6.855 mmol

moles acetic acid reacted = 6.855 mmol

molarity of origjnal acetic acid solution = 6.855 mmol/50 ml = 0.14 M

B)

initial pH = -log[H+] = 2.88

[H+] = [A-] = 1.32 x 10^-3 M

Ka = [H+][A-]/[HA]

     = (1.32 x 10^-3)^2/0.14

     = 1.24 x 10^-5 is the experimental Ka

C) this needs the graph

D) % error = (1.8 x 10^-5 - 1.24 x 10^-5) x 100/1.8 x 10^-5 = 31.11%

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