What is the pH of a solution prepared by adding .490g of ammonium iodide (NH4I) to 155 ml of water? Kb of NH3 is 1.8 *10^-5. (Hint: NH4I immediately dissociates, so you are starting with .490g NH4+)
Kb = [NH3]^2 / [NH4+] = x/0.022=1.8*10^-5, x=3.96*10^-7
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