Question

Calculate the pH at 25 degrees C of 211 mL of a buffer solution that is 0.230 M NH4Cl and 0.230 M NH3 before and after the addition of 1.6 mL of 6.0 M HNO3. The pKa for NH4+ is 9.75.

Answer 1

Before addition:

pH= pKa + log [NH3]/[NH4+]

pH= 9.75 + log 0.23/0.23= 9.75

After addition:

The acid HNO3 will react with NH3 to form NH4+.

mol of acid= 6M x 1.6x10^-3L = 9.6x10^-3 mol

mol NH4+= mol NH3= 0.23M x 211x10^-3L= 0.04853 mol

final mol NH4+= 0.04853 mol + 9.6x10^-3 mol= 0.05813 mol

final mol NH3= 0.04853 mol - 9.6x10^-3 mol= 0.03893 mol

[NH4+]final= 0.05813mol/212.6x10^-3L= 0.273M

[NH3]final= 0.03893mol/212.6x10^-3L= 0.183M

pH= 9.75 + log 0.183/0.273= 9.58