posted by Cait on .
Determine the pH of 263 ml of solution which has [NH 4I] = 0.300 M. Kb = 1.74 × 10-5 for NH3(aq)
NH4+ + H2O ==> H3O+ + NH3
Ka = (Kw/Kb) = (H3O+)(NH3)/(NH4+)
Set up an ICE chart and substitute into the Ka expression. Solve for (H3O+) and convert to pH. You don't need the 263 mL.