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Topic: Calculate Kb for the ascorbate ion  (Read 7174 times)

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Offline alpro

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Calculate Kb for the ascorbate ion
« on: December 19, 2010, 04:24:17 PM »
Sodium ascorbate, the sodium salt of ascorbic acid, is used as an antioxidant in food products. A 0.15 mol/L solution of the ascorbate ion, HC6H6O6^-, has a pH of 8.65. Calculate Kb for the ascorbate ion.
The answer is 1.3 x 10^-10, I just need the steps, kinda confused atm
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Offline alpro

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Re: Calculate Kb for the ascorbate ion
« Reply #1 on: December 19, 2010, 05:03:09 PM »
Got my answer, Thanks for everyone who might have Tried ;)

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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