What is difference between $\ce{H+}$ and a proton?
2 Answers
There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics. A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton.
That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).
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5$\begingroup$ Yes, and also H+ makes me think more about the solvated proton. For instance, in chemical reactions where what shows up in equations as H+ is really H3O+, etc. Using the word proton emphasizes the physics over chemistry, thus it would be odd to refer to proton NMR as H+ NMR, again, that would be 'correct' but confusing. $\endgroup$ Commented Sep 14, 2015 at 5:18
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$\begingroup$ This answer is good and valid in the narrow (but common) case where $\ce{H}$ refers to $_1^{1}\ce{H}$ only (with $\ce{D}$ equalling $^2_1\ce{H}$ and $\ce{T}$ for $^3_1\ce{H}$). This means that in terms of molar mass, there is a difference between the case just outlined and the concept of $\ce{H}$ for hydrogen as an element with different isotopes. $\endgroup$– TAR86Commented Dec 19, 2017 at 17:20
In gaseous/plasmatic phase, there is no difference - $\mathrm{p}$ and $\ce{H+}$ are synonyms for a proton.
The former ( p, proton ) is more often used by physicists in subatomic particle context, the latter by chemists in hydrogen properties/behaviour context.
In polar solvents like water or liquid ammonia, "naked" protons nor electrons cannot exist, but they are solvated, forming molecular ion or reacting with solution components.
$\ce{H+}$, frequently used in electrochemistry and ion equations, is the implied shortcut for hydrated proton aka hydronium $\ce{H3O+}$.
Strong enough acids form solid hydronium salts like e.g. hydronium perchlorate $\ce{[H3O+][ClO4-]}$.
There are consider also larger scale hydronium hydration structures
$\ce{H9O4+ = H3O+ \cdot 3 H2O}$
$\ce{H3O+ } \cdot 6\ \ce{H2O}$
$\ce{H3O+ } \cdot 20\ \ce{H2O}$,
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$\begingroup$ I have not noticed until now the Q and 1st A are quite old, being among active question.:-) $\endgroup$– PoutnikCommented Oct 15, 2019 at 6:39