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I have read in my textbook that an acidic proton is transferred to a base through direct contact. Since acid is ionized in water, wouldn't it be more likely that the surrounding water molecules become protonated to a greater extent than the base (which has less contact with acid)?

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    $\begingroup$ Yes, but then $\ce{H3O+}$ comes into contact with the base, and you end up with the same outcome (the base being protonated), just indirectly. $\endgroup$ Jul 4, 2021 at 23:18

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[OP] Since acid is ionized in water ...

Weak acids are largely protonated, so they still can act as a acid. Strong acids, however, are fully ionized (deprotonated) in water, and can't act as an acid anymore. The protons of strong acids are transfered to water, yielding hydronium ions, which in turn can react as a acid.

[OP] ... wouldn't it be more likely that the surrounding water molecules become protonated to a greater extent than the base (which has less contact with acid)?

The water molecules are able to act as bases, so you can have an acid:base reaction with water. It is true that water is usually present in higher concentration than everything else (otherwise, we would not consider water to be the solvent). For that reason, the most likely encounter is between two water molecules, followed by water with a different species (hydronium ion, hydroxide ion, other acid or base), followed by encounters by non-water species.

[OP] I have read in my textbook that an acidic proton is transferred to a base through direct contact.

This is correct in the sense that an isolated proton does not exist in aqueous solution. So the proton will only dissociate from an acid if it almost immediately associates with a base (this could be water, hydroxide, or another base). Where the proton ends up (or where is predominantly is attached) depends on the $\mathrm{p}K_\mathrm{a}$ values of the conjugate acids of the available bases.

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  • $\begingroup$ Thank you for your response! Regarding your last statement, why would the location of the proton depend on the pKa values of the conjugate acids of the available bases? Would the strongest base not get protonated to the fullest extent (as described by its Kb value) then the next strongest base and so on? $\endgroup$
    – Justin
    Jul 5, 2021 at 0:16
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    $\begingroup$ @Justin It is an equilibrium. So if the strengths are similar, it is near 50:50, and if they are very different, it is almost 100:0, but not quite. It also depends on the concentration of the species. $\endgroup$
    – Karsten
    Jul 5, 2021 at 1:23
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The proton transfer is like this:

$$\ce{HA + B <=> A- + BH+}$$

Where

  • $\ce{HA}$ is any acid, including $\ce{H3O+}$ and $\ce{H2O}$
  • $\ce{B}$ is any base, including $\ce{H2O}$ and $\ce{OH-}$

Note that formally neutral $\ce{HA}$ and $\ce{B}$ may be either neutral molecules or ions.

Concentrations of respective acidic and basic forms are mutually linked by their own equilibrium constants for their water interaction, called acidity resp. basicity constant:

$$K_\mathrm{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

$$K_\mathrm{b}=\frac{\ce{[BH+][OH-]}}{\ce{[B]}}$$

and by water auto-dissociation constant:

$$K_\mathrm{w}=\ce{[H3O+][OH-]}$$

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