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If 97.7 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, what is the molarity of silver ion in the original solution?

Answer :

Final answer:

To find the molarity of silver ion in the original 97.7 ml solution that yielded 0.326 g of AgCl precipitate, calculate the moles of AgCl from its mass and then divide by the solution volume in liters, resulting in a molarity of approximately 0.0232 M.

Explanation:

If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, the molarity of silver ion in the original solution can be determined following these steps:

Identify the precipitate formed which is likely AgCl (silver chloride), due to the reaction between silver nitrate (AgNO₃) and potassium chloride (KCl).

Calculate the moles of the precipitate using its mass and molar mass. The molar mass of AgCl is approximately 143.32 g/mol. Therefore, moles of AgCl = mass / molar mass = 0.326 g / 143.32 g/mol, giving approximately 0.00227 moles of AgCl.

Since the reaction of AgNO₃ with KCl produces AgCl in a 1:1 molar ratio, the moles of Ag⁺ in the original solution are the same as the moles of AgCl precipitated, which is 0.00227 moles.

Finally, to find the molarity of Ag⁺ in the original solution, divide the moles of Ag⁺ by the volume of the solution in liters. The volume of the solution is given as 97.7 mL, which is 0.0977 liters. Therefore, the molarity of Ag⁺ = 0.00227 moles / 0.0977 liters, which equals approximately 0.0232 M.

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Rewritten by : Jeany

The complete balanced chemical reaction is written as:



AgNO3 + KCl ---> AgCl
+ KNO3



where AgCl is our
precipitate





So calculating for moles
of AgCl produced: MM AgCl = 143.5 g/mol



moles AgCl = 0.326 g /
(143.5 g/mol) = 2.27 x 10^-3 mol



we see that there is 1
mole of Ag per 1 mole of AgCl so:



moles Ag = 2.27 x 10^-3
mol





The molarity is simply
the ratio of number of moles over volume in Liters, therefore:



Molarity = 2.27 x 10^-3
mol / 0.0977 L



Molarity = 0.0233 M