Why is the test for sodium as dihydrogen antimonate so specific? I have been attending lectures of salt analysis in my college. We are at group 6 cations or also known as group zero cations, namely $\ce{NH4^+}$, $\ce{K^+}$, $\ce{Na^+}$, $\ce{Mg^2+}$.
The lecturer wrote an equation:
$$\ce{NaCl + KH2SbO4 -&gt; NaH2SbO4 (s) v  + KCl}$$
Now, I know that the electrochemical series follows the order $\ce{Li}$, $\ce{K}$, $\ce{Ba}$, $\ce{Sr}$, $\ce{Ca}$, $\ce{Na}$, $\dots$.
I would like to know why this reaction occurs so spontaneously that it is included as one of the lab tests. Sodium chloride is completely neutral whereas potassium dihydrogen antimonate is basic.
It has no (direct) relation to the electrochemical series nor to acido-basic properties.
It is not a true chemical reaction, rather just crystallization of the least soluble salt from the available ions.
There are no distinguished molecules $\ce{NaCl}$, $\ce{KH2SbO4}$, $\ce{NaH2SbO4}$, $\ce{KCl}$ in the solution, but mutually independent hydrated ions $\ce{Na+}$,  $\ce{Cl-}$, $\ce{K+}$, $\ce{H2SbO4-}$ instead (plus some minor hydrolysis products). Which salt precipitates first depends on their solubility products. $\ce{NaH2SbO4}$ has smaller solubility product than $\ce{KH2SbO4}$.
Similar things would happen if you mix solutions of potassium  chloride and sodium nitrate ( chlorate, perchlorate ). If original solutions are concentrated enough, and especially if you cool the mixture below ambient temperature, there will precipitate the respective potassium salt with the oxoanion.
Another case is the $\ce{NaCl}$ solution  with dissolved $\ce{NH3}$ And $\ce{ CO2}$, with ions $\ce{Na+, NH4+, Cl^-, HCO3-}$, leading to precipitation of $\ce{NaHCO3}$ and solution of $\ce{NH4Cl}$. It is used commercially in the Solvay process
producing sodium carbonate from sodium chloride.