Question

In a solution, the concentration of $\mathrm{KI}$ is $0.01 \mathrm{M}$ and the concentration of $\mathrm{KCl}$ is $0.1 \mathrm{M}$. What is the concentration of $\left[I^{-}\right]$ when $\mathrm{AgCl}$ starts precipitating?

$$ \left[K_{sp}(AgI)=1.5 \times 10^{-16}, K_{sp}(AgCl)=1.8 \times 10^{-10}\right] $$

A $3.5 \times 10^{-7}$

B $6.1 \times 10^{-8}$

C $2.2 \times 10^{-7}$

D $8.3 \times 10^{-8}$

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Answer

Let's solve for the concentration of $ \left[ \mathrm{I}^{-} \right] $ when the precipitation of $\mathrm{AgCl}$ begins:

Given:

  • Concentration of $\mathrm{KI}$: $0.01 \mathrm{M}$
  • Concentration of $\mathrm{KCl}$: $0.1 \mathrm{M}$
  • Solubility product for $\mathrm{AgI}$: $ K_{sp}(\mathrm{AgI}) = 1.5 \times 10^{-16}$
  • Solubility product for $\mathrm{AgCl}$: $ K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$

Step-by-Step :

  1. When the precipitation of $\mathrm{AgCl}$ begins, the ionic product is equal to its solubility product.

  2. For $\mathrm{AgCl}$: $$ K_{sp}(\mathrm{AgCl}) = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] $$

  3. Substituting the given values: $$ 1.8 \times 10^{-10} = [\mathrm{Ag}^{+}] \times 0.1 \mathrm{M} $$

  4. Solving for $[\mathrm{Ag}^{+}]$: $$ [\mathrm{Ag}^{+}] = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \mathrm{M} $$

  5. Now, using the value of $[\mathrm{Ag}^{+}]$, we will find the concentration of $[\mathrm{I}^{-}]$ using the solubility product of $\mathrm{AgI}$: $$ K_{sp}(\mathrm{AgI}) = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] $$

  6. Substituting the values: $$ 1.5 \times 10^{-16} = 1.8 \times 10^{-9} \times [\mathrm{I}^{-}] $$

  7. Solving for $[\mathrm{I}^{-}]$: $$ [\mathrm{I}^{-}] = \frac{1.5 \times 10^{-16}}{1.8 \times 10^{-9}} \approx 8.3 \times 10^{-8} \mathrm{M} $$

Final Answer:

Thus, the concentration of $[\mathrm{I}^{-}]$ when $\mathrm{AgCl}$ begins to precipitate is $8.3 \times 10^{-8}$.

The correct answer is D.


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