Question

Calculate the pH of 100.0 mL of a buffer that is 0.070 M NH4Cl and 0.155 M NH3 before and after the edition of 1.00 mL of 5.90 M HNO3.

Answer 1

You missed the value of Kb of NH₃, or pKb, which in this case is 4.74 (pKb), now write the overall equation and use the hendersson hasselbach equation:

NH₃ + H₂O ---------> NH₄⁺ + OH^-

pOH = pKb + log ([NH₄⁺] / [NH₃])

pOH = 4.74 + log (0.070/0.155)

pOH = 4.39

pH = 14-4.39 = 9.61

After the addition of acid, the concentration of NH₄⁺ is increase while the NH₃ is decreased. so:

moles of NH3 = 0.155 * 0.100 = 0.0155 moles

moles of NH4Cl = 0.070 * 0.1 = 0.007 moles

moles of HNO3 = 5.90 * 0.001 = 0.0059 moles

then:

pOH = 4.74 + log (0.007+0.0059 / 0.0155-0.0059)

pOH = 4.87

pH = 14-4.87 = 9.13

Hope this helps