[Sis290025]

What is the pH of a solution after 0.0600 mol of NaOH is added to a buffer formed by dissolving 0.3500 moles of ammonium chloride (NH4Cl) and 0.9500 moles of ammonia (NH3) in enough water to produce 0.4300 liters of solution? Kb(NH3)= 1.8x10-5

NaOH + NH4Cl --> NH3 + NaCl

Find the final mol. of each substance.

-------NaOH------------------NH4Cl---------------NH3
start---0.06 mol--------------0.35 mol-----------0.95 mol
delta---(-)0.06---------------(-)0.06-------------+0.06
final----0.06 mol-------------0.29-----------------1.01 mol

Find M of acid and conjugate base:

NH4Cl = 0.29 mol/0.43 L = 0.6744186 M
NH3 = 1.01 mol/0.43 L = 2.348837 M

K_a = 1.0E-14/1.8E-5 = 5.555E-10

pH = pK_a + log[base]/[acid]

pH = -log(5.555E-10) + log(2.3488/0.6744186) = 9.80

Thank you.

[Borek]

Seems OK, although I haven't checked very thoroughly. Just the final NaOH should be 0, but it doesn't change anything.