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Chapter 16: Problem 71

What is the \(\mathrm{I}^{-}\) concentration just as \(\mathrm{AgCl}\) begins to precipitate when \(1.0 \mathrm{M} \mathrm{AgNO}_{3}\) is slowly added to a solution containing \(0.020 \mathrm{M} \mathrm{Cl}^{-}\) and \(0.020 M \mathrm{I}^{-}\)

Short Answer

Expert verified
Answer: The concentration of I⁻ just as AgCl starts to precipitate is approximately 9.22 × 10^(-9) M.

Step by step solution

01

Write the solubility product expressions for AgCl and AgI

The solubility product expressions are given as follows: For AgCl: \[ K_{sp}^{AgCl} = [Ag^+][Cl^-] \] For AgI: \[ K_{sp}^{AgI} = [Ag^+][I^-] \] The given values of Ksp for AgCl and AgI are: \( K_{sp}^{AgCl} = 1.8 × 10^{-10} \) and \( K_{sp}^{AgI} = 8.3 × 10^{-17} \)
02

Write the initial concentrations of Cl⁻ and I⁻

The initial concentrations of Cl⁻ and I⁻ ions are: [Cl⁻] = 0.020 M [I⁻] = 0.020 M
03

Calculate the concentration of Ag⁺ when AgCl starts to precipitate

When AgCl starts to precipitate, the product of [Ag⁺] and [Cl⁻] will be equal to its Ksp value. Thus, we can write: \[ [Ag^+] × [Cl^-] = K_{sp}^{AgCl} \] Let the concentration of Ag⁺ be x and [Cl⁻] = 0.020 M. Therefore, \[ x × (0.020) = 1.8 × 10^{-10} \] Now let's solve for x: \[ x = \frac{1.8 × 10^{-10}}{0.020} = 9 × 10^{-9} M \] The concentration of Ag⁺ just as AgCl starts to precipitate is 9 × 10^(-9) M.
04

Calculate the concentration of I⁻ when AgCl starts to precipitate

Now that we have the concentration of Ag⁺, we can calculate the concentration of I⁻ when AgCl starts to precipitate. Using the solubility product expression for AgI, we have: \[ [Ag^+] × [I^-] = K_{sp}^{AgI} \] \[ (9 × 10^{-9}) × [I^-] = 8.3 × 10^{-17} \] Now, let's solve for [I⁻]: \[ [I^-] = \frac{8.3 × 10^{-17}}{9 × 10^{-9}} = 9.22 × 10^{-9} M \] So, the concentration of I⁻ just as AgCl starts to precipitate is approximately 9.22 × 10^(-9) M.

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