# A solution is a mixture of #"0.05M NaCl"# and #"0.05M NaI"#.The concentration of iodide ion in the solution when #"AgCl"# just starts precipitating is equal to?

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#K_("sp AgCl") = 1 * 10^(-10)# #"M"^2#

#K_("sp AgI") = 4* 10^(-16)# #"M"^2#

After you mix the two solutions, the resulting solution will contain two anions that can combine with the silver(I) cations to form an insoluble precipitate, the chloride anions and the iodide anions, respectively.

For silver chloride, the solubility equilibrium looks like this

By definition, the solubility product constant for this equilibrium is equal to

In your case, you know that the solution contains

This means that the minimum concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to

This time, the solubility product constant is equal to

This means that the minimum concentration of silver(I) cations that will cause the silver iodide to precipitate is equal to

Notice that you have

This tells you that as you start to add silver(I) cations to the solution, the silver iodide will precipitate first, consuming some of the iodide ions present in the solution in the process.

This tells you that when silver chloride precipitate, the solution will not contain

because some of the iodide ions already precipitated as silver iodide. In order for the silver(I) cations to precipitate the silver chloride when

you need a concentration of silver(I) cations equal to

present in the solution. This means that at that point, the corresponding concentration of iodide anions must be at a maximum--which corresponds to the solubility equilibrium condition--value of

The answer is rounded to one significant figure.

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To find the concentration of iodide ion when AgCl just starts precipitating, we use the concept of ionic equilibrium. AgCl precipitates when the concentration of Cl- ions exceeds the solubility product constant (Ksp) for AgCl. The Ksp expression for AgCl is [Ag+][Cl-]. Since AgCl and NaCl will both dissociate to give Cl- ions, we can write the expression as [Ag+] [Cl-] = Ksp. At the point of precipitation, all the Cl- ions will come from NaCl, so [Cl-] = 0.05 M. Now, to find the concentration of iodide ions, we set up a similar expression using the Ksp of AgI, [Ag+][I-] = Ksp. Since AgI does not precipitate yet, all Ag+ ions come from NaI, so [Ag+] = 0.05 M. Rearranging the equation gives [I-] = Ksp / [Ag+]. Substituting the known values, [I-] = (8.5 × 10^−17) / (0.05) = 1.7 × 10^−15 M. Therefore, the concentration of iodide ions in the solution when AgCl just starts precipitating is 1.7 × 10^−15 M.

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To determine the concentration of iodide ion in the solution when AgCl just starts precipitating, we need to consider the solubility product constant (Ksp) for AgCl and the reaction that occurs:

AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)

Since AgCl just starts precipitating, the concentration of Ag⁺ and Cl⁻ ions must be equal to the solubility product constant for AgCl:

Ksp = [Ag⁺] * [Cl⁻]

Given that the solution contains both NaCl and NaI, and their concentrations are equal (0.05 M), the concentration of Cl⁻ ions is also 0.05 M.

Now, we need to determine the concentration of Ag⁺ ions when AgCl just starts precipitating. Since AgCl is in equilibrium with Ag⁺ ions, the concentration of Ag⁺ ions will be equal to the solubility of AgCl, which is the square root of the solubility product constant:

[Ag⁺] = √(Ksp / [Cl⁻])

Substituting the known values:

[Ag⁺] = √(1.77 × 10^(-10) / 0.05)

Now, we can calculate [Ag⁺]:

[Ag⁺] ≈ √(3.54 × 10^(-9))

[Ag⁺] ≈ 1.88 × 10^(-4) M

Since NaI dissociates into Na⁺ and I⁻ ions, and their concentration is the same (0.05 M), the concentration of I⁻ ions is also 0.05 M.

Therefore, the concentration of iodide ion in the solution when AgCl just starts precipitating is 0.05 M.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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