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Step-by-Step Solution
Step 1: Identify the Relevant Solubility Product Expressions
We are given two sparingly soluble salts:
(i) Silver chloride ($\text{AgCl}$) with $K_{sp}(\text{AgCl}) = 1.7 \times 10^{-10}\,\text{M}^2$.
(ii) Silver chromate ($\text{Ag}_2\text{CrO}_4$) with $K_{sp}(\text{Ag}_2\text{CrO}_4) = 1.9 \times 10^{-12}\,\text{M}^3$.
The solubility product expressions are:
1) $K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-]$.
2) $K_{sp}(\text{Ag}_2\text{CrO}_4) = [\text{Ag}^+]^2\,[\text{CrO}_4^{2-}]$.
Step 2: Write Down the Given Ion Concentrations
From the question:
• $[\text{Cl}^-] = 0.1\,\text{M}$.
• $[\text{CrO}_4^{2-}] = 0.001\,\text{M} = 10^{-3}\,\text{M}$.
The volume change on adding $\text{AgNO}_3$ is assumed negligible, so these concentrations remain effectively the same.
Step 3: Find the $\text{Ag}^+$ Concentration Required to Precipitate AgCl First
For $\text{AgCl}$ to start precipitating:
$$
K_{sp}(\text{AgCl}) = [\text{Ag}^+]_{\text{AgCl}} \times [\text{Cl}^-].
$$
Hence,
$$
[\text{Ag}^+]_{\text{AgCl}}
= \frac{K_{sp}(\text{AgCl})}{[\text{Cl}^-]}
= \frac{1.7 \times 10^{-10}}{0.1}
= 1.7 \times 10^{-9}\,\text{M}.
$$
Step 4: Find the $\text{Ag}^+$ Concentration Required to Precipitate Ag₂CrO₄
For $\text{Ag}_2\text{CrO}_4$:
$$
K_{sp}(\text{Ag}_2\text{CrO}_4)
= [\text{Ag}^+]^2 \,[\text{CrO}_4^{2-}].
$$
Thus,
$$
[\text{Ag}^+]_{\text{Ag}_2\text{CrO}_4}
= \sqrt{ \frac{K_{sp}(\text{Ag}_2\text{CrO}_4)}{[\text{CrO}_4^{2-}]} }
= \sqrt{ \frac{1.9 \times 10^{-12}}{10^{-3}} }
= \sqrt{1.9 \times 10^{-9}}.
$$
Numerically,
$$
\sqrt{1.9 \times 10^{-9}}
\approx \sqrt{1.9}\times 10^{-4.5}
\approx 1.378 \times 3.16 \times 10^{-5}
\approx 4.36 \times 10^{-5}\,\text{M}.
$$
(Depending on the approximation, some texts represent it as $\sqrt{19} \times 10^{-4}\,\text{M}$, which also indicates an order of magnitude of about $10^{-4}\,\text{M}$.)
Step 5: Compare the Required $\text{Ag}^+$ Concentrations
• To start precipitating $\text{AgCl}$: $[\text{Ag}^+] \approx 1.7 \times 10^{-9}\,\text{M}$.
• To start precipitating $\text{Ag}_2\text{CrO}_4$: $[\text{Ag}^+] \approx 4.36 \times 10^{-5}\,\text{M}$.
Clearly, $1.7 \times 10^{-9}\,\text{M}$ is much smaller than $4.36 \times 10^{-5}\,\text{M}$. Therefore, $\text{AgCl}$ will precipitate at a lower $\text{Ag}^+$ concentration than $\text{Ag}_2\text{CrO}_4$.
Step 6: Conclude Which Salt Precipitates First
Since $\text{AgCl}$ requires a significantly lower concentration of $\text{Ag}^+$ to start its precipitation, $\text{AgCl}$ will precipitate first as we gradually add $\text{AgNO}_3$.
Final Answer
AgCl will precipitate first as the amount of Ag+ needed for its precipitation is lower than that required for Ag2CrO4.
