Question

Vinegar Titration Curve / / 14.00 13.00 12.00 11.00 10.00 9.00 8.00 E 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 25.00 30.00 35.00 15.00 20.00 Volume NaOH (mL)

Given that the pKa for the half equivalence point is = 5.5, and the pKa of Acetic acid is 4.76, what is the percent error?

Is 4.76 the accepted value?
and 5.5 the experimental?

and so...

|4.76-5.5| = -0.74
---------    ----- = 0.15546 ×100% = 15.55%
   |4.76|         |4.76|

Please verify I am correct.

Answer 1

The equivalence point locate at the point where very small addition of titrant let  a very rapid rise in the pH. Graphically, it is the point on the titration curve where the slope, ΔpH/ΔV, takes change from positive to negative. As per your question , titration curve for the titration of Acetic acid by NaOH, a weak acid and strong base. The weak acid dissociates into hydrogen ion and conjugate base of acid and changes the appearance of titration curve.

For the dissociation of any weak acid,

HA(aq) → H+ (aq) + A– (aq)

pH= pKa + log ([A-]/ [HA])

The Henderson-Hasselbach equation help to calculate the pKa of an acid. At the equivalence point, the volume of base added is just enough to exactly neutralize all of acid. At one-half of this volume of added base, called half equivalence point,  The Theoritical pH= pKa at that point as half of the acid reacts to form A- , the conc. of A- and HA at half-equivalence point fall same. Thus, at half-equivalence point, the pH is equal to the pKa. so, value is 5.5.

Thus , 4.76 is the theoretical equivalence point , Whereas 5.5 is the experimental half equivalence point.

% Error = [(5.5-4.76)/4.76]*100 = 15.54%