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Shouldn't 3-bromopropanoic acid have the least pKb due to the reduced inductive effect and electronegativity, leaving the carboxylate anion with more electric charge on it? Also, what will be the decreasing order of pKb?

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  • $\begingroup$ Least pKb implies that it is the most basic of all four, which can be predicted using the inductive effect logic as you figured out. $\endgroup$
    – Arishta
    Jan 31 '17 at 15:27
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The correct order of PKb is

fluoroacetic acid> 3-fluoropropanoic acid>3-chloropropanoic acid>3-bromopropanoic acid.

To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton.

Halogens are electron withdrawing groups. The halogen which will withdraw the most electron density will stabilise the anion the best. Fluorine, being the most electronegative withdraws the most electron density thus reducing the charge on oxygen and stabilising the anion, therefore having the least pKa and maximum pKb.

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  • $\begingroup$ RE: "To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton." This seems backwards. "The stability of negative charge on carboxylate anion" is essentially the pKb. I think rather you need the notion that pKa + pKb = 14 for aqueous solutions. $\endgroup$
    – MaxW
    Jan 31 '17 at 17:13

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