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The Henderson–Hasselbalch equation can be used to model these equilibria. It is important to maintain this pH of 7.4 to ensure enzymes are able to work optimally. [10] Life threatening Acidosis (a low blood pH resulting in nausea, headaches, and even coma, and convulsions) is due to a lack of functioning of enzymes at a low pH. [10]
For acetic acid, K a = 1.8 x 10 −5, so pK a is 4.7. A higher K a corresponds to a stronger acid (an acid that is more dissociated at equilibrium). The form pK a is often used because it provides a convenient logarithmic scale, where a lower pK a corresponds to a stronger acid.
K 1, K 2 and DIC each have units of a concentration, e.g. mol/L. A Bjerrum plot is obtained by using these three equations to plot these three species against pH = −log 10 [H +] eq, for given K 1, K 2 and DIC. The fractions in these equations give the three species' relative proportions, and so if DIC is unknown, or the actual concentrations ...
For K′ 3 there are three different dissociation constants — there are only three possibilities for which pocket is filled last (I, II or III) — and one state (I–II–III). Even when the microscopic dissociation constant is the same for each individual binding event, the macroscopic outcome (K′ 1, K′ 2 and K′ 3) is not equal. This ...
4] = [H +] T ( 1 + [SO 2− 4] / K * S) −1. However, it is difficult to estimate K * S in seawater, limiting the utility of the otherwise more straightforward free scale. Another scale, known as the seawater scale, often denoted pH SWS, takes account of a further protonation relationship between hydrogen ions and fluoride ions, H + + F − ...
From the titration of protonatable group, one can read the so-called pK a 1 ⁄ 2 which is equal to the pH value where the group is half-protonated (i.e. when 50% such groups would be protonated). The pK a 1 ⁄ 2 is equal to the Henderson–Hasselbalch pK a (pK HH a) if the titration curve follows the Henderson–Hasselbalch equation. [14]
1, then, [CrO 2− 4] = 1 / K 1 K D [H +] . Blue line Chromate and dichromate have equal concentrations. Setting [CrO 2− 4] equal to [Cr 2 O 2− 7] in Eq. 3 gives [CrO 2− 4] = 1 / β 2 [H +] 2 . The predominance diagram is interpreted as follows. The chromate ion is the predominant species in the region to the right of the ...
Buffer capacity falls to 33% of the maximum value at pH = pK a ± 1, to 10% at pH = pK a ± 1.5 and to 1% at pH = pK a ± 2. For this reason the most useful range is approximately pK a ± 1. When choosing a buffer for use at a specific pH, it should have a pK a value as close as possible to that pH. [2]