Chem 1212 Lab Manual -Revised 12/20s REPORT SHEET Determination of the Ka for a

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Chem 1212 Lab Manual -Revised 12/20s REPORT SHEET Determination of the Ka for a Weia Name Adwea Boniu Partner's Name Samn RobinsnIn Except where otherwise noted, 3pts or all bnk e Ka for a Weak Acid Date Section Section coinsn Iastructer Batec ame Sarouh fo Instructor BakeC NaoH Molarity 100 Efficiency of pi meter Since the titration with 1-ml increments is done to give a rough estimate of the equivalence point, you need report on only the results from the more precise 0.5-mL. increment titration. Acetic Asid Literature value for K, of Acetic Acid Literature value for pK, 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 error of K Show both percent error calculations (4 pou 31 4.74 30.5ml 6.30 310) pts.) :34,997 )1.99102-1.80x10 ( .Ty 19-2 Volume of base to reach equivalence point plH at equivalence point Volume of base halfway to equivalence point pHH at halfway to equivalence point 65

Explanation / Answer

Calculation of concentration of acetic acid:

The reaction is-    CH3COOH + NaOH = CH3COONa + H2O

At the equivalence point, no of milimoles of acid = no of milimoles of base

or, V1S1 = V2S2                    (always remember to take the concentration in normality)

where, V1 = volume of acid taken

S1 = concentration of acid (in normality)

V2 = volume of base to reach to the equivalence point

S2 = concentration of base ( in normality)

concentration of base = 0.1035 M = 0.1035 N      (as NaOH is a monoacidic base)

V1S1 = V2S2

or, 10 mL * S1 = 30.5 mL * 0.1035 N

or, S1 = 0.3157 N

concentration of acetic acid = 0.3157 N = 0.3157 M

(normality and molarity of acetic acid is same because acetic acid is a monobasic acid, i.e., 1 mole of acetic acid gives 1 mole of H+ in solution. But if it were dibasic acid, e.g. H2SO4, normality would not be equal to molarity. In that case, molarity = normality/2, because 1 mole of H2SO4 gives two moles of H+ in solution.)

Calculation of concentration of the unknown acid:

[Note that you don't know if the unknown acid is monobasic or not, so in this case here you cannot assume molarity of unknown acid = normality of the unknown acid .]

V1S1 = V2S2

or, 10 mL * S1 = 19.2 mL * 0.1035 N

or, S1 = 0.1987 N

concentration of the unknown acid = 0.1987 N

[If it is not mentioned in the experiment if the unknown acid is monobasic acid or not, take it as a monobasic one, then concentration of unknown = 0.1987 M]

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