At second equivalence point , the species present is C6H6O6 2- mole of Ascorbic
ID: 699748 • Letter: A
Question
At second equivalence point , the species present is C6H6O62-
mole of Ascorbic = (0.945mol/1000ml)×921ml = 0.870345
No of mole of NaOH required = 2×0.870345 = 1.74069
Volume of NaOH required = (1000/1.69mol)×1.74069mol = 1030ml
Dilution of C6H6O22- = 1951ml/921ml =2.1183time
Concentration of C6H6O62- = 0.945M/2.1183 = 0.44611M
C6H6O62- is partly hydrolysed by water
C6H6O62- + H2O --------> HC6H6O6- + OH-
Kb = [OH-][HC6H6O6-]/[C6H6O62-]
Kb = Kw/Ka2
= 1.00 ×10^-14/1.6×10^-12
= 6.25×10^-3
at equillibrium
[OH-] = x
[ HC6H6O6-] = x
[ C6H6O62-] = 0.44611 - x
x^2/(0.44611 - x ) = 6.25 ×10^-3
x = 0.04977
[ OH- ] = 0.04977M
pOH = -log(0.04977)
= 1.30
pH = 14 - 1.30
= 12.70
Explanation / Answer
Consider a the titration of 0.921 L of 0.945 M ascorbic acid (H2CGH60) with 1.69 M NaOH. What is the pH at the second equivalence point of the titration?
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.