Search results
Results from the WOW.Com Content Network
For acid–base reactions, the equivalent weight of an acid or base is the mass which supplies or reacts with one mole of hydrogen cations (H +). For redox reactions, the equivalent weight of each reactant supplies or reacts with one mole of electrons (e −) in a redox reaction. [3]
By this definition, the number of equivalents of a given ion in a solution is equal to the number of moles of that ion multiplied by its valence. For example, consider a solution of 1 mole of NaCl and 1 mole of CaCl 2. The solution has 1 mole or 1 equiv Na +, 1 mole or 2 equiv Ca 2+, and 3 mole or 3 equiv Cl −.
Normality can be used for acid-base titrations. For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be ...
Historically, the mole was defined as the amount of substance in 12 grams of the carbon-12 isotope.As a consequence, the mass of one mole of a chemical compound, in grams, is numerically equal (for all practical purposes) to the mass of one molecule or formula unit of the compound, in daltons, and the molar mass of an isotope in grams per mole is approximately equal to the mass number ...
The solvent (e.g. water) is omitted from the defining expression on the assumption that its concentration is very much greater than the concentration of dissolved acid, [H 2 O] ≫ T A. The equation for mass-balance in hydrogen ions can then be written as T H = [H +] + [A −][H +]/K a − K w / [H +]
It was exactly equal before the redefinition of the mole in 2019, and is now only approximately equal, but the difference is negligible for all practical purposes. Thus, for example, the average mass of a molecule of water is about 18.0153 daltons, and the molar mass of water is about 18.0153 g/mol.
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:
To create the solution, 11.6 g NaCl is placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL. The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore, the molar concentration of water is