This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: a , you must 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 HH 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 E5 pKa = 4.76. pH = pKa – log [(HA) / (A)] pH = 4.76 – log [1.25 / 0.75] pH = 4.76 – 0.222 pH = 4.5 11. Calculate the pH of a buffer solution made from 0.20 M HC 2 H 3 O 2 and 0.050 M C 2 H 3 O 2that has an acid dissociation constant for HC 2 H 3 O 2 of 1.8 x 105 . Solve this problem by plugging the values into the HendersonHasselbalch 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 105 ) + log (0.050 M / 0.20 M) pH =  log (1.8 x 105 ) + log (0.25) pH = 4.7  0.6 pH = 4.1...
View
Full Document
 Spring '10
 WU
 Biochemistry, Plants, 0.050 M, Van der Waals, 0.20 m, Waals

Click to edit the document details