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?

0.0726 M

Explanation:

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