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Unformatted text preview: SA or SB)
do • Good buffer preparation usually tries to produce
a 50% acid (or base) + 50% conjugate.
• Buffer Capacity: moles of SA or SB that are
required to change buffer by one pH unit (+/–)
• See Exercise 8.8 and end of chapter problems
68 and 69 to practice buffer capacity.
68 HA H+ + A–
• In a typical Ka set up, Ka = [H+][A–]/[HA]. So…
the ratio of Ka (HA/A– ) should ~ [H+].
• Most often interested in pH -- could take
the “– log” of both sides and obtain
• – logKa + ( – log(HA/A– ) = – log[H+] or…..
• pKa – log ([HA]/[A–]) = pH. This becomes
pH = pKa + log ( [A–]/[HA]) for weak acids.
For bases, pH = pKa + log (B/BH+) The Henderson–Hasselbalch Equation
• Buffers are solutions that resist changes in pH (you
have plenty in you right now).
• Typically, buffers consist of a weak acid and its
conjugate (salt A–) or a weak base and its conjugate
(salt BH+). We can just add moles of each to a set
volume or titrate a weak acid with a strong base to
make conjugate in the solution.
• With the proper K and concentrations, we can obtain
the pH of a buffered solution. Conversely, we can...
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