20 Structure and Acid Strength

# 20 Structure and Acid Strength - An Acid-Base Equilibrium...

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1 1 An Acid-Base Equilibrium Problem HF is a weak acid. A 0.25 M HF solution has a pH of 1.92. Calculate the pK a of HF. Note: In these acid-base problems, the concentration specified is always the initial concentration not the equilibrium concentration, unless told otherwise. Set up the RICE table. Use the pH to find the amount of acid that ionizes. Solve for K a and then pK a . A) 3.42 B) 4.76 C) 3.22 D) 6.47 E) 4.23 HF + H 2 O H 3 O + + F - pH = - log[H 3 O + ] pK a = - log(K a ) 2 Solution HF is a weak acid. A 0.25 M of HF solution has a pH of 1.92. Calculate the pK a . So K a = (0.012x0.012)/(0.25-0.012) = 6.05 x 10 -4 pK a = 3.22 HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) K a = [HF] [H 3 O + ][F - ] 0.25 M 0 0 -x x x .25-x 10 -1.92 10 -1.92 = 0.012

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2 3 Calculate the pH of a 0.25 M solution of NH 3 given pK b = 4.745. Plan: Recognize as an acid-base equilibrium problem! Construct the RICE table. Write the K b expression. Use K b to calculate [OH - ]. Then calculate pOH from [OH - ]. And calculate pH from the pOH. Possible Answers A) 2.67 B) 11.3 C) 3.45 D) 10.8 4 Solution .25 M 0 0 -x x x .25-x x x  4 3 2 4.745 3 10 0.25 2.12x10 [ ] 2.67 14 2.67 11.3 b NH OH K NH x x x OH pOH pH       Assume 0.25-x ~ 0.25 Remember what x is!
3 5 Content identified on the Course Schedule posted on Bb 26 multiple choice questions, 2 count as bonus questions your score = number correct x 100 / 24 Be sure to bring the following: • Stony Brook ID card, Calculator with spare batteries • 2 or more #2 pencils and a good eraser

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## This note was uploaded on 03/29/2010 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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20 Structure and Acid Strength - An Acid-Base Equilibrium...

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