What is the quantity of material of NH4Cl that should be added to a solution 0.1 of NH3 to form 1L of a solution with ph 8? Kb = 1, 8x10 ^-5

Answer 1

pKa for NH3 = about 9.74 but should do that conversion yourself and use something other than my estimate. Also, I assume you mean 0.1M in NH3.
pH = pKa + log (NH3/NH4Cl)
8.0 = 9.74(or whatever) + log 0.1/(NH4Cl)
Solve for (NH4Cl) and convert to grams for a liter of solutiion.

Answer 2

Thanks :)