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The Henderson–Hasselbalch equation relates the pH of a solution containing a mixture of the two components to the acid dissociation constant, K a of the acid, and the concentrations of the species in solution. [6] Simulated titration of an acidified solution of a weak acid (pK a = 4.7) with alkali
At half-neutralization the ratio [A −] / [HA] = 1; since log(1) = 0, the pH at half-neutralization is numerically equal to pK a. Conversely, when pH = pK a, the concentration of HA is equal to the concentration of A −. The buffer region extends over the approximate range pK a ± 2. Buffering is weak outside the range pK a ± 1.
For example, if a macromolecule M has three binding sites, K′ 1 describes a ligand being bound to any of the three binding sites. In this example, K′ 2 describes two molecules being bound and K′ 3 three molecules being bound to the macromolecule. The microscopic or individual dissociation constant describes the equilibrium of ligands ...
V eq is the volume of titrant (ml) consumed by the crude oil sample and 1 ml of spiking solution at the equivalent point, b eq is the volume of titrant (ml) consumed by 1 ml of spiking solution at the equivalent point, 56.1 g/mol is the molecular weight of KOH, W oil is the mass of the sample in grams. The normality (N) of titrant is calculated as:
The K factor or characterization factor is defined from Rankine boiling temperature °R=1.8Tb[k] and relative to water density ρ at 60°F: . K(UOP) = / The K factor is a systematic way of classifying a crude oil according to its paraffinic, naphthenic, intermediate or aromatic nature. 12.5 or higher indicate a crude oil of predominantly paraffinic constituents, while 10 or lower indicate a ...
The equation for mass-balance in hydrogen ions can then be written as T H = [H +] + [A −][H +]/K a − K w / [H +] Titration curves for addition of a strong base to a weak acid with pK a of 4.85. The curves are labelled with the concentration of the acid. where K w represents the self-dissociation constant of water.
For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, because of the existence of isotopes, molar masses are used instead in calculating the mass ratio.
Its conjugate base is the acetate ion with K b = 10 −14 /K a = 5.7 x 10 −10 (from the relationship K a × K b = 10 −14), which certainly does not correspond to a strong base. The conjugate of a weak acid is often a weak base and vice versa.