1. Write the reaction

Silver nitrate and potassium iodide react as:

$\mathrm{Ag^+ + I^- \rightarrow AgI(s)}$

Spectator ions are $\mathrm{K^+}$ and $\mathrm{NO_3^-}$.

2. Calculate initial moles

For $\mathrm{AgNO_3}$:

$V = 25\,\text{mL} = 0.025\,\text{L}, \quad M = 1.0$

$n(\mathrm{Ag^+}) = 1.0 \times 0.025 = 0.025\,\text{mol}$

For $\mathrm{KI}$:

$V = 25\,\text{mL} = 0.025\,\text{L}, \quad M = 1.05$

$n(\mathrm{I^-}) = 1.05 \times 0.025 = 0.02625\,\text{mol}$

Also,

$n(\mathrm{K^+}) = 0.02625\,\text{mol}, \qquad n(\mathrm{NO_3^-}) = 0.025\,\text{mol}$

3. Find limiting reagent

Reaction is $1:1$ between $\mathrm{Ag^+}$ and $\mathrm{I^-}$.

Since

$0.025 < 0.02625$

$\mathrm{Ag^+}$ is limiting, so all $\mathrm{Ag^+}$ precipitates as $\mathrm{AgI}$.

Excess iodide left:

$n(\mathrm{I^-\,left}) = 0.02625 - 0.025 = 0.00125\,\text{mol}$

Total volume after mixing:

$25 + 25 = 50\,\text{mL} = 0.050\,\text{L}$

So bulk concentrations are approximately:

$[\mathrm{I^-}] \approx \frac{0.00125}{0.050} = 0.025\,\text{M}$

$[\mathrm{K^+}] = \frac{0.02625}{0.050} = 0.525\,\text{M}$

$[\mathrm{NO_3^-}] = \frac{0.025}{0.050} = 0.50\,\text{M}$

Clearly $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ are present in large amounts.

4. Check ions present in very small quantity

Even though almost all $\mathrm{Ag^+}$ precipitates, because solid $\mathrm{AgI}$ is present, there will be a tiny amount of both $\mathrm{Ag^+}$ and $\mathrm{I^-}$ in solution governed by solubility equilibrium:

$\mathrm{AgI(s) \rightleftharpoons Ag^+ + I^-}$

with

$K_{sp} = [\mathrm{Ag^+}][\mathrm{I^-}]$

Since excess iodide is present, $[\mathrm{I^-}]$ is finite and $[\mathrm{Ag^+}]$ becomes extremely small. Also, compared with spectator ions, the dissolved iodide associated with the sparingly soluble salt is also the ion involved in the very low-solubility equilibrium. Thus the ions occurring in very small amount due to the precipitation equilibrium are $\mathrm{Ag^+}$ and $\mathrm{I^-}$.

Hence, among the options, the intended answer is:

$\boxed{\text{C: } \mathrm{Ag^+} \text{ and } \mathrm{I^-} \text{ both}}$

5. Compare with stored correct answer

Stored correct answer: C

Derived answer: C

They match.