Why is the enthalpy of solution for NaCl 3.88 and not 4.1 kJ/mol? Based on Hess' law, the molar enthalpy of solution is equal to the sum of the enthalpies of formation of products minus reactants. In the case of dissolving table salt ($\ce{NaCl}$) in water, those reactants are $\ce{NaCl (s), Na+ (aq)}$, and $\ce{Cl- (aq)}$, with enthalpies for formation of $-411.2$, $-239.7$, and $-167.4\ \mathrm{kJ/mol}$, respectively.

So, based on Hess' law, the enthalpy of solution should be $4.1\ \mathrm{kJ/mol}$. However, the actual enthalpy of solution in this case is $3.88\ \mathrm{kJ/mol}$ according to multiple sources. The discrepancy appears to be that the enthalpies of formation for $\ce{Na+}$ and $\ce{Cl-}$ don't exactly equal the combined enthalpy of formation for $\ce{NaCl (aq)}$.

I am at loss for why there is this small offset. Is it due to only partial dissociation of $\ce{NaCl (aq)}$ while in solution? And, if so, wouldn't that have a larger effect? Or is the effect slight because there are some inter-molecular forces at play that only minimally affect the total enthalpy of solution? It could be that my numbers are wrong, but a cursory search on the internet matches the values for $\ce{Na+}$ and $\ce{Cl-}$ in the textbook I have on hand.
Your approach is correct. The small discrepancy of the results when using parameter values from different sources could be caused by inconsistent standard states (temperature and pressure), or simply by measurement uncertainty (in this case, the resulting difference would not be considered significant).

By way of comparison, I found the following parameter values for a temperature of $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ and a pressure of $p=1\ \mathrm{bar}=100\ \mathrm{kPa}$.


Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})=-411.2\ \mathrm{kJ\ mol^{-1}}$ [1]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})=-240.1\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})=-167.2\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of solution $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})=3.88\ \mathrm{kJ\ mol^{-1}}$ [3]


Calculating the standard molar enthalpy of solution from the given values for the standard molar enthalpy of formation yields

$$\begin{align}
\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})&amp;=\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})+\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})-\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})\\[6pt]
&amp;=-240.1\ \mathrm{kJ\ mol^{-1}}-167.2\ \mathrm{kJ\ mol^{-1}}+411.2\ \mathrm{kJ\ mol^{-1}}\\[6pt]
&amp;=3.9\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

Considering the number of significant digits, there is no significant difference between the calculated result of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.9\ \mathrm{kJ\ mol^{-1}}$ and the corresponding literature value of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.88\ \mathrm{kJ\ mol^{-1}}$.



References

[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[2] “Thermodynamic Properties of Aqueous Ions”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[3] “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.