SHOW ALL WORK! Also help with graphs I created information is provided. Here are
ID: 705846 • Letter: S
Question
SHOW ALL WORK! Also help with graphs I created information is provided.
Here are the graphs clearer (copy and paste)
file:///C:/Users/1234567/Documents/Acetic%20acid%20with%20NaOH.pdf
file:///C:/Users/1234567/Documents/Unknown%20with%20NaOH.pdf
Calculations:
a) Experimental pKa of Acetic Acid
b) Percent error of pKa
c) Experimental Ka of Acetic Acid
Information for graphs:
Graphs I created:
Ka for a Weak Acid Section Determination of the REPORT SHEET Name Partner's Name Date Instructor 91.8 NaOH Molarity 0:1024m.) with 1-mL increments is done to give a rough estimate of the equivalence Efficiency of pH meter Since the titration ?you need report on only the results from the more precise increment titration. Literature value for Ka of Acetic Acid Literature value for pKs of Acetic Acid Volume of base to reach equivalence point pH at equivalence point Volume of base halfway to equivalence point pH at halfway to equivalence point Experimental pK, of Acetic Acid Percent error of pK Experimental K, of Acetic Acid Percent eror of Ka Show both percent error calculations (4peu: 435 63 6.92 Lo 915 S pts.) 4.2s ps) Unknown code Volume of base to reach equivalence point pH at equivalence point Volume of base halfway to equivalence point +-25 75Explanation / Answer
To find the value of pKa, we must locate the volume of the equivalence point (20.63 mL) followed by this, calculate the average volume of equivalence point (20.63 / 2 = 10.32 mL), at that volume the pH read is equal to pKa of the acid (pH = pKa = 4.8), then we calculate the error between the read pKa and the literature:
% e = (4.75 - 4.8) * 100 / 4.75 = 1.05% Next we calculate the Ka: Ka = Antilog (-pKa) = Antilog (-4.8) = 1.58 * 10 ^ -5 and finally his error: % e = (1.58 - 1.8) * 10 ^ -5 * 100 / 1.8 * 10 ^ -5 = 12.22% Answers: (a) pKa = 4.8 (b)% e = 1.05% (c) Ka = 1.58 * 10 ^ -5 (d)% e = 12.22%
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