Calculate the pH of 100.0 mL of a buffer that is 0.0550 M NH4Cl and 0.150 M NH3 before and after the addition of 1.00 mL of 5.20 M HNO3. _____ = pH before _____= pH after

Answer 1

pOH = pKb + log [NH₄Cl]/[NH₃]

pKb = -log Kb and looking up the Kb for NH3 I find a value of 1.8x10^-5

pK = 4.74

Before addition of anything:

pOH = 4.74 + log (0.0550/0.150) = 4.74 - 0.44

pOH = 4.30

pH = 9.70

After addition of 1.00 ml of 5.20 M HNO₃:

HNO₃ will react with NH₃ to produce NH₄⁺

NH₃ + H⁺ ==> NH₄⁺

mols NH₃ in 100 mls = 0.1 L x 0.150 mol/L = 0.0150 mols NH₃

mols H⁺ added = 0.001 L x 5.20 mol/L = 0.0052 mols H⁺

mols NH₃ after addition of HCl = 0.0150 - 0.0052 = 0.0098 mols NH₃

mols NH₄⁺ formed = 0.0052

final mols NH⁴ = 0.1 L x 0.0550 mol/L = 0.0055 mol + 0.0052 mol = 0.0107 mols NH₄⁺

Assuming no volume change (1ml added to 100 ml, we can neglect this change in volume)

pOH = 4.74 + log (0.0107 / 0.0098) = 4.74 +0.038

pOH = 4.78

pH = 9.22