Question

A 1.40 g sample of silver nitrate is dissolved in sufficient water to make 125 ml of solution. To this solution is added 5.00 ml of 1.50 M hydrochloric acid, and a precipitate forms. Find the concentration of silver ions remaining in solution?

Answer 1

By calculating the moles of AgNO3 and HCl, and considering the reaction stoichiometry and solubility, the remaining concentration of Ag+ ions in the solution is approximately 5.7 x 10^-5 M.

Step-by-step explanation:

To find the concentration of silver ions remaining in solution after the reaction between silver nitrate and hydrochloric acid, we need to perform a series of calculations. The reaction is as follows:

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Firstly, calculate the moles of AgNO3 used:

(1.40 g AgNO3) / (169.87 g/mol) = 0.00824 mol AgNO3

Since the reaction ratio is 1:1, the moles of Ag+ initially present in the new solution will also be 0.00824 mol. Now, calculate the moles of HCl:

(5.00 mL HCl) × (1 L / 1000 mL) × (1.50 mol/L) = 0.0075 mol HCl

Since AgCl is insoluble, all the available Cl− from HCl will precipitate with Ag+ until one reactant is used up. Since HCl is in excess, the remaining concentration of Ag+ will be:

(0.00824 mol Ag+ - 0.0075 mol Ag+) / (0.130 L) ≈ 5.7 x 10−5 M