Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. For AgCl, K_sp = 1.8 *10¯¹⁰ A) 1.8 * 10¯¹⁰ mol Ag⁺ B) 9.0 * 10¯¹⁰ mol Ag⁺ C) 9.0 * 10¯⁹ mol Ag⁺ D) 6.7 *10¯⁸ mol Ag⁺ E) 1.3 *10¯⁵ mol Ag⁺

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Quizplus · Accepted Answer

From the given equation: K_sp = 1.8 F1F1F1S1F1F1F10 10¯¹⁰ we can find the solubility product of AgCl, which is: K_sp = 1.8 × 10^-10 When KCl is added, it will react with Ag+ to form AgCl and K+. This will increase the concentration of Cl- in the solution, causing more AgCl to dissolve until equilibrium is reestablished. Since the volume does not change, we can assume that the concentration of Ag+ remains constant at equilibrium. Let x be the additional amount of AgCl that dissolves. Then, the concentration of Cl- in the solution is 0.200/(1000+x) mol/L. Since Ag+ and Cl- ions have a 1:1 stoichiometry, the concentration of Ag+ ions is also 0.200/(1000+x) mol/L. The total concentration of Ag+ ions in the solution is the sum of the initial concentration and the additional amount that dissolves: [Ag+] = 1.8 × 10^-10 mol/L + 0.200/(1000+x) mol/L Since [Ag+] = [Cl-], the total concentration of Ag+ and Cl- ions (i.e. AgCl) in the solution is: [AgCl] = 2[Ag+] = 2[Cl-] = 2 × 0.200/(1000+x) mol/L The molar solubility of AgCl is defined as the amount of AgCl that dissolves in one liter of solution, which is given by the expression: s = x/(1000+x) mol/L Using the solubility product equation for AgCl, we can express x in terms of s: K_sp = ([Ag+][Cl-])^2 1.8 × 10^-10 = (0.200/(1000+x))^2 x = 1.346 × 10^-8 mol/L Substituting this value of x into the expression for the molar solubility of AgCl, we get: s = 1.346 × 10^-8 / (1000 + 1.346 × 10^-8) = 1.345 × 10^-11 mol/L Since AgCl dissociates into Ag+ and Cl- ions in a 1:1 ratio, the concentration of Ag+ ions in the solution is also 1.345 × 10^-11 mol/L. Therefore, the number of moles of Ag+ ions in the solution is: 1.345 × 10^-11 mol/L × 1.0 L = 1.345 × 10^-11 mol Since the only ion that contains Ag in the solution is Ag+ ion, this is also the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. Converting to scientific notation and rounding to one decimal place gives the final answer of 6.7 × 10^-12 mol Ag⁺. Therefore, the correct answer is D.

Assuming That the Total Volume Does Not Change After 0