Search results
Results from the WOW.Com Content Network
The distribution constant (or partition ratio) (K D) is the equilibrium constant for the distribution of an analyte in two immiscible solvents. [1] [2] [3]In chromatography, for a particular solvent, it is equal to the ratio of its molar concentration in the stationary phase to its molar concentration in the mobile phase, also approximating the ratio of the solubility of the solvent in each phase.
Substance Formula 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 60 °C 70 °C 80 °C 90 °C 100 °C Barium acetate: Ba(C 2 H 3 O 2) 2: 58.8: 62: 72: 75: 78.5: 77: 75
For example, potassium oxide is about 83% potassium by weight, while potassium chloride is only 52%. Potassium chloride provides less potassium than an equal amount of potassium oxide. Thus, if a fertilizer is 30% potassium chloride by weight, its standard potassium rating, based on potassium oxide, would be only 18.8%.
[10]: 280–4 Hence, a single experiment can be used to measure the logarithms of the partition coefficient (log P) giving the distribution of molecules that are primarily neutral in charge, as well as the distribution coefficient (log D) of all forms of the molecule over a pH range, e.g., between 2 and 12.
However, since water is in vast excess, the concentration of water is usually assumed to be constant and is omitted from equilibrium constant expressions. Often, the metal and the ligand are in competition for protons. [note 4] For the equilibrium p M + q L + r H ⇌ M p L q H r. a stability constant can be defined as follows: [28] [29]
The following chart shows the solubility of various ionic compounds in water at 1 atm pressure and room temperature (approx. 25 °C, 298.15 K). "Soluble" means the ionic compound doesn't precipitate, while "slightly soluble" and "insoluble" mean that a solid will precipitate; "slightly soluble" compounds like calcium sulfate may require heat to precipitate.
The value of the equilibrium constant for the formation of a 1:1 complex, such as a host-guest species, may be calculated with a dedicated spreadsheet application, Bindfit: [4] In this case step 2 can be performed with a non-iterative procedure and the pre-programmed routine Solver can be used for step 3.
However, the equilibrium constant for the loss of two protons applies equally well to the equilibrium [M(H 2 O) n] z+ - 2 H + ⇌ [MO(H 2 O) n-2] (z-2)+ + H 2 O. because the concentration of water is assumed to be constant. This applies in general: any equilibrium constant is equally valid for a product with an oxide ion as for the product with ...