Solid ammonium bromide is slowly added to 175 mL of a 0.153 M silver nitrite solution until the concentration of bromide ion is 0.0628 M. The percent of silver ion remaining in solution is ____ %.

Answer 1

Ammonium bromide react with Silver Nitrite to form AgBr which is very insoluble.  The solubility

product constant Ksp is 5.0 x 10^-13.

The Ksp expression is then [Ag¹⁺] [Br^1-] = 5.0 x 10^-13.

After adding the ammonium bromide, the concentration of bromide is said to be 0.0628 M.

Substitute 0.0628 into the Ksp equation to get[Ag¹⁺] [0.0628] = 5.0 x 10^{-1 5.2}and solve for the silver ion

concentration.

[Ag¹⁺] =  7.96 x 10^-12 M. Since the initial molarity was 0.153 M, the percent remaining is 7.96 x 10^-12 / 0.153 x 100 = 5.20 x 10^-9