Acetic acid CH 3 COOH is a weak acid with K a 18 x 10 5 A buffer is prepared by

Acetic acid ch 3 cooh is a weak acid with k a 18 x 10

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17. Acetic acid, CH3COOH, is a weak acid with Ka = 1.8 x 10-5. A bufferisprepared by adding 0.60 moles CH3COOH, and 0.60 moles CH3COO-Na+(sodium acetate) to a total of 1.00 L of solution. What is the pH of this bufferafter 0.20 moles of HCl is added? A. 3.77B. 5.05C. 4.15D. 5.45E. 4.44
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A- + H3O+ ←→ HA + HReactions with strong acids go to completion. Therefore, use the Large K Rule.A- + H3O+ ←→ HA + HInitial 0.60 0.20 0.60Change -0.20 -0.20 +0.20 +0.20Equilibrium 0.40 0 0.80Henderson-Hasselbalch equation:pH = pKa + log(base/acid)pH = (-log(1.8 x 10-5)) + (log(0.40/0.80)) = 4.44CHEM162-2014 CHAPTER 17 2 O 2 O 16Kimmel’s notes:Given: Weak acid, conjugate base, strong base, given initial concentrations and Ka; find pH.0.010 mol NaOH was added to 1.00 liter of a buffer containing 0.50M CH3COOH and0.50M CH3COONa. What is the pH? K . -5 -10 is a CHEM162-2014 CHAPTER 17 17 Z&Z 39a. ET: Given weak base, conjugate acid, strong acid, K b - Find pH.
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Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL ofthe following buffered solution. ) 4 4 4 4 4 -5 -10 4 4 CHEM162-2014 CHAPTER 17 18
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Effect of Buffer Concentration on pH What is the pH of a buffer of 5.00 M HA and 5.00M NaA? K a = 1.8x10 -5 HA + H 2 O ←→ H 3 O + + AHA( aq) + H 2 O( l ) ←→ H 3 O + (aq) + A - (aq) Initial 5.00 0 5.00 Change Equilibrium HA(aq) + H 2 O( l ) ←→ H 3 O + (aq) + A - (aq) Initial 5.00 0 5.00 Change -X +X +X Equilibrium 5.00 - X +X 5.00+X Small K rule followed by Henderson-Hasselbalch equation. HA(aq) + H 2 O( l ) ←→ H 3 O + (aq) + A - (aq) Initial 5.00 0 5.00 Change -X +X +X Equilibrium 5.00 +X 5.00 pH = pK a + log([B]/[A]) pH = -log(1.8x10 -5 ) + log([5.00]/[5.00]) pH = 4.74
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