Question

What is the concentration of Ag⁺ ion in solution after 100.0mL of 0.150 M AgNO₃ is added to 20.0 ml of 1.00 M NaBr? Then find the concentration of Br-, Na+, and No3- ion as well

***** I figured out that the mass of solid AgBr produces 2.82 grams, but don't know how to apply that to figure out the answer for the question above

Answer 1

[AgNO3] =[Ag^+] = [NO3^-] = 100 mL x 0.150 mol / 1000 mL => 0.015 mol

[NaBr] = [Na^+] = [Br^-] = 20 mL x 1.00 mol / 1000 mL => 0.020 mol

After mixing total volume of solution = 100 + 20 = 120 mL or 0.120 L

AgNO3 (aq) + NaBr (aq) ----> AgBr (s) + NaNO3 (aq)

0.015 mol       0.020 mol     0.015 mol    0.015 mol    + 0.005 mol NaBr unreacted

[AgBr] = 0.015 mol x 187.77 g/mol =>2.82 g

There will be no free Ag^+ after reaction. it will be form of AgBr solid and its mass = 2.82 g

[Br^-] = 0.005 mol / 0.120 L => 0.0416 M

[Na^+] = 0.020 mol / 0.120 L => 0.1667 M

[NO3^-] = 0.015 mol / 0.120 L => 0.1250 M