8 and end of chapter problems see 68 and 69 to

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Unformatted text preview: SA or SB) do • Good buffer preparation usually tries to produce Good a 50% acid (or base) + 50% conjugate. 50% • Buffer Capacity: moles of SA or SB that are moles required to change buffer by one pH unit (+/–) required • See Exercise 8.8 and end of chapter problems See 68 and 69 to practice buffer capacity. 68 HA H+ + A– • In a typical Ka set up, Ka = [H+][A–]/[HA]. So… In/[HA]. the ratio of Ka (HA/A– ) should ~ [H+]. • Most often interested in pH -- could take Most the “– log” of both sides and obtain the • – logKa + ( – log(HA/A– ) = – log[H+] or….. log or….. • • • pKa – log ([HA]/[A–]) = pH. This becomes pH = pKa + log ( [A–]/[HA]) for weak acids. for For bases, pH = pKa + log (B/BH+) The Henderson–Hasselbalch Equation The • Buffers are solutions that resist changes in pH (you Buffers have plenty in you right now). have • Typically, buffers consist of a weak acid and its Typically, conjugate (salt A–) or a weak base and its conjugate conjugate or (salt BH+). We can just add moles of each to a set (salt ). volume or titrate a weak acid with a strong base to make conjugate in the solution. make • With the proper K and concentrations, we can obtain With the pH of a buffered solution. Conversely, we can...
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This document was uploaded on 03/28/2014.

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