Volume at equivalence point? Volume at half-equivalence point? pKa and Ka? REPOR

Question

Volume at equivalence point? Volume at half-equivalence point? pKa and Ka?
REPORT SHEET EXPERIMENT Determination of the Dissociation Constant of a Weak Acid 25 B. Standardization of Sodium Hydroxide (NaOH) Solution Trial I Trial 2 Trial 3 Mass of bottle + KHP Mass of bottle Mass of KHP used 21.0?'mL 41.9°'ml Final buret reading Initial buret reading mL of NaOH used 0.0921 nity of NaoH0.0471 0.09? Averagermolanty (show calculations and standard deviation) Standard deviation (see Experiment 8) 341 Copyright 0 2015 Pearson Education, Inc.

Explanation / Answer

Recall that at the equivalence point, the volume of base added is just enough to exactly neutralize all of the acid. At one-half of this volume (of added base), it is called the half-equivalence point, i.e. enough base has been added to neutralize half of the acid.

Now observe the three columns carefully,

In first determination at volume =13.8ml of NaOH there is a sharp change in the pH value, similar sharp changes are seen at 13.2ml and 13.6 ml for second and third determination.

So averaging the volume at equivalence point for the three determinations, the volume is     (13.8+13.2+13.6)/3 = 13.53ml.... (Answer)

Volume at half equivalence point is just the half of the volume at equivalence point, thus

Volume at half equivalence point = 13.53/2=6.76ml. .... (Answer)

pKa is the pH at half equivalence point, so here pKa= 4.47. .... (Answer)

(Note: as the readings are taken after 1 ml addition we will have to approximate half equivalent point as 7ml instead of 6.67ml)

Ka=antilog (-pKa) =antilog (-4.47) =0.0114 =11.4*10^-3.... (Answer)

Hope this Helps!!

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