25\,mL of 1\,M AgNO3 is added to 25\,mL of 1.05\,M KI solution. The ion(s) present in very small quantity in the…

🧠 The precipitate locks both $\ce{Ag+}$ and $\ce{I-}$ After mixing, the strongly insoluble $\ce{AgI}$ forms. $\ce{K+}$ and $\ce{NO3-}$ are spectators — they remain in full concentration. The two ions that build the precipitate ($\ce{Ag+}$ and $\ce{I-}$) are present only in trace amounts in the supernatant solution because of $K_\text{sp}$.

🗺️ Per-option check

• mmol $\ce{AgNO3} = 25 \times 1 = 25$.
• mmol $\ce{KI} = 25 \times 1.05 = 26.25$.
• $\ce{Ag+}$ and $\ce{I-}$ react in $1:1$, leaving only $1.25\,\text{mmol}$ of $\ce{I-}$ in $50\,\text{mL}$ — and even that gets tied down by the small $K_\text{sp}$ of $\ce{AgI}$.

(a) "$\ce{I-}$ only" — wrong; $\ce{Ag+}$ is also extremely small (it's the limiting one). (b) "$\ce{K+}$ only" — wrong; $\ce{K+}$ is a bulk spectator at $\sim 0.5\,\text{M}$. (c) "$\ce{NO3-}$ only" — wrong; same reason as $\ce{K+}$. (d) Both $\ce{Ag+}$ and $\ce{I-}$ are tied up by the precipitate, so both are in very small amounts.

⚠️ Common mistake Picking (a) because $\ce{Ag+}$ is the limiting reagent. Yes, $\ce{Ag+}$ runs out — but the leftover $\ce{I-}$ is also pulled down by $K_\text{sp}$ of $\ce{AgI}$. So both end up in tiny equilibrium amounts.

$\boxed{\text{Answer: (d)}}$