If 37.1 mL AgNO3 solution reacts with excess potassium chloride solution to yield 1.56 g of AgCl precipitate, what is the molarity of silver ion in the original solution? AgCl = 143.3 g/mol Enter your answer in decimal format with three decimal places and no units.

Molarity of silver ion: 0.296 M

Explanation:

Reaction:

AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻

From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.

First, let's calculate the number of moles of AgCl produced:

1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.

The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:

moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol

Molarity of silver ion = 0.296 M