JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Equilibrium

Mixing gives $25\times 1=25$ mmol $\mathrm{AgNO}{}_3$ and $25\times 1.05=26.25$ mmol $\mathrm{KI}$; silver nitrate is the limiting reagent and nearly all the silver is removed as the highly insoluble precipitate $\mathrm{AgI}$. The solution is then dominated by the spectator ions $\mathrm{K}{}^{+}$ (26.25 mmol) and $\mathrm{NO}{}_3{}^{-}$ (25 mmol). The silver ion remaining is fixed by the extremely small solubility product of $\mathrm{AgI}$, and the iodide left is only the small 1.25 mmol excess, whose free concentration is further suppressed by the common-ion effect. Hence both $\mathrm{Ag}{}^{+}$ and $\mathrm{I}{}^{-}$ are present in very small quantity.