In my teacher's power-point states that we must consider that inductive effect in $CH_{3}COOH$ exists because of the electron deficiency in $C$ (partial positive charge $\delta^{+}$) as a result of the electronegativity of the double bonded $O$. As you may notice in the figure 1 although the single bonded $O$ is more electronegative than $C$, the electron density of the single bonded $O$ moves to $C$ in order to "neutralize" the $\delta^{+}$, leaving $H$ less attached to the molecule.
Why do we consider that double bonded $O$ establishes that $\delta^{+}$ on our $C$ (and creates an electron withdrawing inductive effect $-I$) and not the single bonded $O$? Is this because of the $\pi$ double bond? Can you tell me why?
Figure 1: Inductive effect created by the electron deficiency of $C$. As we can see the $H$ is now less attached to our molecule
Now supose we have the following carboxylic acids:
$CH_{3}COOH$, pka = $4.80$
$FCH_{2}COOH$, pka = $2.66$
$O_{2}^{-}CCH_{2}COOH$, pka = $5.69$
$O_{2}^{-}C(CH_{2})_{4}COOH$, pka = $5.41$
For me it makes sense that $FCH_{2}COOH$ is more acidic than $CH_{3}COOH$, because $F$ is an electronegative element and he creates a partial negative charge moving de electron density to him, leaving our $H$ in the carboxyl group less attached and easier to remove.
On the other hand the fact $O_{2}^{-}CCH_{2}COOH$ is less acidic than $O_{2}^{-}C(CH_{2})_{4}COOH$ for me it's less intuitive. In my teacher's power-points states that $O_{2}^{-}$ linked to an alkyl chain will create an electron donating inductive effect $+I$. Can you explain to me why? Isn't there a withdrawing resonance effect $-R$? If $O$ is an electronegative element why doesn't it create an $-I$ effect instead of $+I$? Beacause if that was the case adding carbons $(CH_{2})_{4}$ in the chain would reduce the the $-I$ effect and the $O_{2}^{-}C(CH_{2})_{4}COOH$ would be less acidic than $O_{2}^{-}CCH_{2}COOH$, and that isn't the case.