Question

You add 56.3 grams of silver nitrite (AgNO$_3$; MW=169.87 g/mol) to 100.0 mL of water. To this solution, you add 80.00 mL of 1.00 M HCl. 1. Prove that precipitate will form and calculate the amount of silver chloride solid (in grams) that would precipitate from the above solution given the $K_{sp}$ of AgCl is 1.6x10$^{-10}$. (MW of silver chloride 143.32 g/mol).

          You add 56.3 grams of silver nitrite (AgNO$_3$; MW=169.87 g/mol) to 100.0 mL of water. To this solution, you add 80.00 mL of 1.00 M HCl.
1. Prove that precipitate will form and calculate the amount of silver chloride solid (in grams) that would precipitate from the above solution given the $K_{sp}$ of AgCl is 1.6x10$^{-10}$. (MW of silver chloride 143.32 g/mol).

You add 56.3 grams of silver nitrite (AgNO3; MW=169.87 g/mol) to 100.0 mL of water. To this solution, you add 80.00 mL of 1.00 M HCl.
1. Prove that precipitate will form and calculate the amount of silver chloride solid (in grams) that would precipitate from the above solution given the Ksp of AgCl is 1.6x10^-10. (MW of silver chloride 143.32 g/mol).

Added by Melissa M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Hitendra Singh

01/03/2022

Step 1

Moles of AgNO3 = mass / molar mass Moles of AgNO3 = 56.3 g / 169.87 g/mol Moles of AgNO3 = 0.331 moles Show more…

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You add 56.3 grams of silver nitrite (AgNO3; MW = 169.87 g/mol) to 100.0 mL of water. To this solution, you add 80.00 mL of 1.00 M HCl. Prove that a precipitate will form and calculate the amount of silver chloride solid (in grams) that would precipitate from the above solution given the Ksp of AgCl is 1.6 x 10^(-10) (MW of silver chloride 143.32 g/mol).