0.15 mol/L solution of the ascorbate ion, HC6H6O6¯, has a pH of 8.65. Calculate the Kb...

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0.15 mol/L solution of the ascorbate ion, HC6H6O6¯, has a pH of 8.65. Calculate the Kb...

0.15 mol/L solution of the ascorbate ion, HC6H6O6¯, has a pH of 8.65. Calculate the Kb for the ascorbate ion.

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Answer 1

Answer #1

Solution :

C6H7O6(aq)+H2O(l)↔C6H8O6(aq)+OH-(aq)

pOH= 14-pH
= 14-8.65
= 5.35

-lg[OH-]= 5.35
[OH-]= 1/(10^5.35)
= 4.47x10^-6M

Kb= [C6H8O6][OH-]/[C6H7O6]
At equilibrium, [C6H8O6]= [OH-]
Therefore, [OH-]^2= (Kb)([C6H7O6])
Kb= ([OH-]^2)/[C6H7O6]
= ((4.47x10^-6)^2)/0.15
= 1.33x10^-10

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Answer 2

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0.15 mol/L solution of the ascorbate ion, HC6H6O6¯, has a pH of 8.65. Calculate the Kb...