Chapter 15 Lecture 2

# Chapter 15 Lecture 2 - pH of 11. CH 3 NH 2 + H 2 O CH 3 NH...

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Chapter 15 Lecture 2 Here is a problem where you calculate the Ka of the acid. Helen has a headache. She makes up a 0.02 M solution of acetylsalicylic acid (asprin). Not wanting to upset her stomach she measures the pH to be 2.5. What is the pKa for asprin? 1) HA + H 2 O H 3 O + + A - Assume water ionization is not a factor Define x HA H 3 O + A - Initial Change Equilibrium ] [ ] ][ [ 3 HA A O H Ka - + =

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Weak base equilibria are similar to weak acids but use the base reaction and the Kb. Ammonia, a simple base NH 3 + H 2 O NH 4 + + OH - Assume water is not a factor Define x Find the pH of a 0.15 M solution of ammonia. Assume water ionization is not a factor NH 3 NH 4 + OH - Initial Change Equilibrium ] [ ] ][ [ 3 4 NH OH NH Kb - + =
Find the concentration of a methylamine solution that has a

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Unformatted text preview: pH of 11. CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH-Assume water is not a factor Define x CH 3 NH 2 CH 3 NH 3 + OH-Initial Change Equilibrium ] [ ] ][ [ 2 3 3 3 NH CH OH NH CH Kb-+ = Polyprotic acids H 2 CO 3 + H 2 O H 3 O + + HCO 3-Ka1 = 4.2 X 10-7 HCO 3-+ H 2 O H 3 O + + CO 3 2-Ka2 = 4.7 X 10-11 What is the pH of a 0.1 M carbonic acid solution? Biggest K wins so can ignore the second reaction. The pH is figured out by the x method. - Note also that H 3 O + HCO 3 With this approximation you can reintroduce the second equilibrium and find CO 3 2-...
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## This note was uploaded on 04/16/2008 for the course CHEM 1123 taught by Professor Paul during the Fall '06 term at Arkansas.

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Chapter 15 Lecture 2 - pH of 11. CH 3 NH 2 + H 2 O CH 3 NH...

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