Answer:
[NH3] = 0.270M
[NH4Cl] = 0.327M
Explanation:
The HNO3 will react with the weak base, NH3, as follows:
HNO₃ + NH₃ → NH₄⁺ + NO₃⁻
Initial moles of each specie of the buffer:
NH3 = NH4⁺ 0.210L * (0.300mol/L) = 0.063moles
The moles added of HNO3 = Additional moles of NH4Cl and the moles substracted of NH3:
0.001L * (6mol / L) = 0.006 moles.
After the addition:
Moles NH3 = 0.063mol - 0.006mol = 0.057moles
Moles NH4Cl = 0.063mol + 0.006mol = 0.0069moles
And their concentrations are:
[NH3] = 0.057moles / 0.211L = 0.270M
[NH4Cl] = 0.069moles / 0.211L = 0.327M
