CHEM 404 Lecture Notes - Lecture 8: Sodium Acetate, Conjugate Acid, Acid Dissociation Constant
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Log(4. 2 x 10-13) = 12. 38: see attached, to prepare a 0. 1 m buffer for a ph of 4. 5 i would choose the acid acetic acid. 4. 74 which is very close to my desired ph of 4. 5. [acetic acid]+[sodium acetate-] = 0. 1 (0. 56)x [sodium acetate]= 0. 1 m. Moles of acetate= 0. 064 x 1 = 0. 064. Moles of acetic acid = 0. 936 x 1 = 0. 936. [sodium acetate]=0. 064 m (0. 064 m/ 0. 1 l) x (1 m/ 82. 0343 g) = 0. 0078 g sodium acetate. [acetic acid]= 0. 936 m (0. 936 m/ 0. 1 l) x (1 m/ 60. 052 g) = 0. 1559 g acetic acid: my molar ratios support the acid to conjugate base concentrations in my buffer system. I have a larger concentration of acetic acid than sodium acetate. The ratio of acetic acid to sodium acetate is 14. 625 times more. This occurs because i am trying to lower the ph from. Therefore, i need more acid than base in my solution.
