Problem: The value of Ka at 25 °C for ascorbic acid (HC6H7O6 , vitamin C), is 8.0 × 10–5. What is pKb of the ascorbate ion?A. 4.1 B. 9.9 C. 10.3 D. 6.1 E. 8.9

FREE Expert Solution

We are being asked to find pKb  of the ascorbate ion. Recall that pKb can be calculate from Kb using the following equation:

pKb=-log Kb


However, we're given the Ka for ascorbic acid (HC6H7O6 , vitamin C). Ascorbate ion is the conjugate base of ascorbic acid.

   HC6H7O6    +      H2O          C6H7O6-       +        H3O+ 
     (weak acid)                    (base)               (conjugate base)              (conjugate acid) 


Hence, the Kb oC6H7O6- can be calculated from the Ka of HC6H7O6.  Ka and Kb are connected by the autoionization constant of water (Kw) in the following equation:

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Problem Details

The value of Ka at 25 °C for ascorbic acid (HC6H7O6 , vitamin C), is 8.0 × 10–5. What is pKb of the ascorbate ion?

A. 4.1 

B. 9.9 

C. 10.3 

D. 6.1 

E. 8.9

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