How many grams of NH4Cl do I have to add to 500 mL of 0.100 M NH4OH to have a pH of 9.40?

Question

Kb= 1.8 x 10^-5
[OH-]= 2.51 x^-5 M

Answer 1

Kb = (NH4^+)(OH^-)/(NH4OH)
Solve for (NH4^+).
Then M NH4^+ = moles NH4^+/L
You know M and L, solve for moles.
Then moles NH4Cl = grams NH4Cl/molar mass MH4Cl. Solve for grams.