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AP Chemistry Acid Bases Problem

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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+)

Oct 23rd, 2017

Kb = [NH3]^2 / [NH4+] = x/0.022=1.8*10^-5, x=3.96*10^-7

[NH4I]=0.49g/(145g/mol)/0.144L=0.022

so [H+]=10^-14/(3.96*10^-7)=2.53*10^-8

so pH=-log[H+]=8

Let me know if you have questoins

Apr 10th, 2015

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Oct 23rd, 2017
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Oct 23rd, 2017
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