Study Questions for Week 1 - Answers

Study Questions for Week 1 - Answers - 2 H 3 O 2-that has...

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BISC 330L Spring 2011 Study Questions for Week 1 (Tabancay) SOLUTIONS FOR QUESTIONS #9-12 9. A buffer is prepared containing 1.00 M acetic acid and 1.00 M sodium acetate. What is its pH? We must know the pK a of acetic acid. Don’t worry, most of the time you will be provided with the pK a . If the problem provides the K a , you must then convert it to the pK a . The K a of acetic acid is 1.77 x 10¯ 5 pK a = - log K a = - log 1.77 x 10¯ 5 = 4.752 Next, we simply insert the appropriate values into the H-H equation: pH = 4.752 + log (1.00 / 1.00) Since the log of 1 is zero, we have pH = 4.752 10. Find the pH of 1.25 M acetic acid and 0.75 M potassium acetate. Acetic acid Ka = 1.74 E-5 pKa = 4.76. pH = pKa – log [(HA) / (A)] pH = 4.76 – log [1.25 / 0.75] pH = 4.76 – 0.222 pH = 4.5 also… pH = pKa + log [(A) / (HA)] pH = 4.76 + log [0.75 / 1.25] pH = 4.76 + (-0.222) pH = 4.5
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BISC 330L Spring 2011 Study Questions for Week 1 (Tabancay) 11. Calculate the pH of a buffer solution made from 0.20 M HC 2 H 3 O 2 and 0.050 M C
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Unformatted text preview: 2 H 3 O 2-that has an acid dissociation constant for HC 2 H 3 O 2 of 1.8 x 10-5 . Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base. pH = pK a + log ([A-]/[HA]) pH = pK a + log ([C 2 H 3 O 2-] / [HC 2 H 3 O 2 ]) pH = - log (1.8 x 10-5 ) + log (0.050 M / 0.20 M) pH = - log (1.8 x 10-5 ) + log (0.25) pH = 4.7 - 0.6 pH = 4.1 12. 1. Calculate the concentration of benzoic acid and benzoate in a solution whose pH = 5.2. The pKa of benzoic acid is 4.20 and the solution was prepared to be 0.005 M benzoic acid. pH = pKa + log[A-/HA] 5.20 = 4.20 + log[A-/HA] 1.0 = log[A-/HA] 10 = [A-]/[HA] says the concentration of benzoate is ten times that of benzoic acid or the fractional amount is: [A-]/([HA] + [A-]) or 1/11 [HA] and 10/11 [A-]. The fractional amount of [HA] is: 0.005 M x 1/11 = .0004545 M (4.5 x 10-4 M) The fractional amount of [A-] is: 0.005 M x 10/11 = .004545 M (4.5 x 10-3 M)...
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This note was uploaded on 02/15/2012 for the course BISC 330L taught by Professor Petruska,tower during the Spring '07 term at USC.

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Study Questions for Week 1 - Answers - 2 H 3 O 2-that has...

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