20 Structure and Acid Strength

# 20 Structure and Acid Strength - An Equilibrium Problem HF...

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1 1 An 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

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2 3 Equilibrium Problem Calculate the pH of a 0.25 M solution of NH 3 given pK b = 4.745. Plan: Recognize as an 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 What determines which of two acids will be the stronger, i.e. which will have the larger K a ?

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20 Structure and Acid Strength - An Equilibrium Problem HF...

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