If a 30.00mL aliquot of a 0.15 M NH4Cl solution is added to 20.00mL of a 0.18 M NH4NO3 solution, what is the concentration of nitrate ion in the mixture?
(NO3^-) = 0.18M x (20/50) = ?
To find the concentration of nitrate ion in the mixture, we need to consider the initial concentration of nitrate ion in the 20.00 mL of the 0.18 M NH4NO3 solution, as well as the dilution caused by adding the 30.00 mL aliquot of the 0.15 M NH4Cl solution.
First, let's calculate the amount of nitrate ion initially present in the 20.00 mL of the 0.18 M NH4NO3 solution:
Amount of nitrate ion = volume (L) x concentration (mol/L)
Amount of nitrate ion = 0.020 L x 0.18 mol/L = 0.0036 mol
Next, let's calculate the concentration of nitrate ion in the 30.00 mL aliquot of the 0.15 M NH4Cl solution:
Amount of nitrate ion = volume (L) x concentration (mol/L)
Amount of nitrate ion = 0.030 L x 0 mol/L (since NH4Cl does not contain nitrate ion) = 0 mol
Now, let's calculate the total volume of the mixture after adding the aliquot:
Total volume = volume of NH4NO3 solution + volume of NH4Cl aliquot
Total volume = 20.00 mL + 30.00 mL = 50.00 mL = 0.050 L
Finally, let's calculate the concentration of nitrate ion in the mixture:
Concentration of nitrate ion = total amount of nitrate ion / total volume
Concentration of nitrate ion = (amount of nitrate ion in NH4NO3 solution + amount of nitrate ion in NH4Cl aliquot) / total volume
Concentration of nitrate ion = (0.0036 mol + 0 mol) / 0.050 L = 0.072 mol/L
Therefore, the concentration of nitrate ion in the mixture is 0.072 M.