A solution is made by dissolving 54.0 g of silver nitrate in enough water to make 350.0 ml. A 10.00 ml portion of this solution is then diluted to a final volume of 250.0 ml. What is the total concentration of ions present in the final solution?

Molar concentration of the final solution is the number of moles of solute (n) divided by the final volume in liters (V). This is the mathematical formula for molarity:

M = n solute / V solution in liters = n silver nitrate / V solution in liters

1) Calculate the number of moles of silver nitrate in the initial solution, dissolving 54.0 g of silver nitrate to make 350.0 ml of solution:

Solute = silver nitrate = AgNO₃

Molar mass of AgNO₃ = 169.87 g/mol

n = mass in grams / molar mass = 54.0 g / 169.87 g/mol = 0.318 mol

2) Calculate the molarity of the initial solution:

M = n / V = 0.318 mol / 0.3500 liter = 0.9086 M

3) Calculate the number of moles of silver nitrate in a 10.0 ml portion of that solution:

M = n / V ⇒n = M × V

n = 0.9086 M × 10.0 ml / (1,000 ml / liter) = 0.009086 moles silver nitrate

4) Molar concentration (molarity) of the final 250.0 ml solution:

M = n / V = 0.009086 mol / 0.2500 liter = 0.0363 M

5) Concentration of ions present in the final solution:

You must assume 100% ionization:

AgNO₃ → Ag⁺ + NO₃⁻

Then, since each mole of AgNO₃ yields two moles of ions, the total concentration of ions present is the double: 2 × 0.0363 M = 0.0726 M