Determine the pH of 263 ml of solution which has [NH 4I] = 0.300 M. Kb = 1.74 × 10-5 for NH3(aq)

Answer 1

I managed to get it :)
though for others who might be looking to solve this for their practice questions

Convert Kb to Ka
Ka = kw/Kb

then just use an ice table with Ka=x^2/ 0.300M-x (x is small)
and just -log[H3O+)

Answer 2

4.9