Mixed 600 cm3 agno3 solution (molar concentration 0.05 mol / dm3) and 400 cm3 NaCL solution

Mixed 600 cm3 agno3 solution (molar concentration 0.05 mol / dm3) and 400 cm3 NaCL solution (molar concentration 0.1 mol / dm3) Calculate the molar concentration of all ions in the solution after separation of the precipitate.

Let’s write the reaction equation:

AgNO3 + NaCl = AgCl (precipitate) + NaNO3.

Let us find the amount of substances n of all reagents: n (AgNO3) = c (AgNO3) * V (AgNO3) = 0.05 * 600 = 30 mol, where c (AgNO3) is the molar concentration of the silver nitrate solution, V (AgNO3) is the volume of the solution silver nitrate.

n (NaCl) = c (NaCl) * V (NaCl) = 0.1 * 400 = 40 mol, where c (NaCl) is the molar concentration of sodium chloride solution, V (NaCl) is the volume of sodium chloride solution.

Thus, we find that sodium chloride was in excess, which means we are calculating according to the lack. Let us assume that when removing the sediment, the volume of the solution V did not change and remained equal to V = V (AgNO3) + V (NaCl) = 600 + 400 = 1000 cm3

We write down the ionic equation Ag + + NO3 – + Na + + Cl- = AgCl (sediment) + Na + + NO3-

Since silver ions are in short supply, they reacted completely and were removed as a precipitate. Chlorine ions reacted with silver ions in an amount of 30 mol, but since they were in excess, they remained n (Cl-) = n (NaCl) – n (AgNO3) = 40 – 30 = 10 mol,

Sodium ions and nitrate ions remained unchanged, the volume of the solution changed.

Let us determine their molar concentrations with:

s (Na +) = n (NaCl) / V = ​​40/1000 = 0.04 mol / cm3

s (NO3-) = n (AgNO3) / V = ​​30/1000 = 0.03 mol / cm3

s (Cl-) = n (Cl -) / V = ​​10/1000 = 0.01 mol / cm3