1.40 g of silver nitrate is dissolved in 125 mL of water. To this solution 5.00 mL of 1.50M hydrochloric acid is added, and a precipitate forms. Find the concentration of silver ions remaining in solution. A) 5.7*10^-3 M B) 6.34*10^-2 M C) 5.77*10^-2 M D) 0.121M E) 5.9*10^-3 M

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Quizplus · Accepted Answer

The concentration of silver ions remaining in solution is calculated by considering the solubility product (Ksp) of silver chloride (AgCl). The Ksp of AgCl is 1.77 x 10^-10. Given that 5.00 mL of 1.50 M HCl is added, the amount of Cl- ions introduced is 0.0075 moles (5.00 mL * 1.50 mol/L * 0.001 L/mL). Since AgCl precipitates, the concentration of Ag+ ions in equilibrium with the precipitated AgCl can be found using the Ksp expression: [Ag+] = Ksp / [Cl-]. Assuming all Cl- ions remain in solution and using the Ksp value, the concentration of Ag+ ions can be calculated, which is approximately 5.9 x 10^-3 M, matching the format of choice A.

140 G of Silver Nitrate Is Dissolved in 125 ML