Given 1 M solutions of acetic acid and sodium acetate, and distilled water, describe the preparation of 1 L of 0.1 M acetate buffer at pH = 5.4. The relevant pKa is 4.76.
I know how to solve this problem but i'm having a hard time converting the moles to liters (like the correct set up too use)
given
pH = 5.4.
pKa is 4.76.
now
pH = pKa +log([A-]/[HA])
[A-] is concentration of base
[HA] is concentration of acid
so
5.4 = 4.76 + log([A-]/[HA])
log([A-]/[HA]) = 0.64
[A-]/[HA] = 4.37
1 L of 0.1 M acetate buffer
here
[A-] is CH3COO-
[HA] is CH3COOH
[CH3COOH] + [CH3COO-] = 0.1M
==>
[CH3COOH] = 0.1 - [CH3COO-]
[CH3COO-]/0.1 - [CH3COO-] = 4.37
solving this gives
[CH3COO-] = 0.08M
[CH3COOH] = 0.1 - 0.08 = 0.02M
1 L 0.1 M acetate buffer is prepared by 0.08 M CH3COO- and 0.02 M CH3COOH with volume of solution 1L
Given 1 M solutions of acetic acid and sodium acetate, and distilled water, describe the preparation...
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