(from Fundamentals of Biochemistry by Voet, 5th ed.)
In this step of glycolysis, I'm not seeing where the $\ce{H+}$ ion on the product side is coming from. It seems to me that the G3P's aldehydic H is replaced by phosphate, and that H is given to $\ce{NAD+}$ to make NADH. So where is the extra $\ce{H+}$ coming from?
Although texts such as Berg et al. tend to refer to inorganic phosphate, $\ce{P_i}$, as orthophosphate ($\ce{PO4^{3-}}$), the term inorganic phosphate is used because in aqueous solution at pH 7.6 several phosphate species exist, the predominant one being $\ce{HPO4^{2-}}$. If this is regarded as $\ce{P_i}$, then it is the source of the $\ce{H^{+}}$, and the equation balances.
Note added by David:
On checking I find, in contrast to Berg, Lehninger’s book defines $\ce{P_i}$ as monohydrogen phosphate ($\ce{HPO4^{2-}}$), and Fersht actually writes the equation of the reaction (16-1) with $\ce{HPO4^{2-}}$ rather than $\ce{P_i}$.
Although an answer that I have contributed an edit to has been, rightly, accepted, I decided to add my own answer to clarify a couple of points.
Chemical equations are written in terms of specific ionic and elemental species and balance elements and charge, whereas biochemical equations are written in terms of reactants that often consist of species in equilibrium with each other and do not balance elements that are assumed fixed, such as hydrogen at constant pH.
One should not, therefore, expect Voet’s equation to balance.
…which answers the question.
It is interesting that the IUPAC document defines Pi as orthophosphate, although this is irrelevant to the question at hand as none of the old papers I have looked at use the abbreviation Pi.
