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This sodium hydroxide solution can be used to measure the equivalent weight of an unknown acid. For example, if it takes 13.20±0.03 cm 3 of the sodium hydroxide solution to neutralise 61.3±0.1 mg of an unknown acid, the equivalent weight of the acid is:
An equivalent (symbol: officially equiv; [1] unofficially but often Eq [2]) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount (typically one mole) of another substance in a given chemical reaction. It is an archaic quantity that was used in chemistry and the biological sciences (see Equivalent weight § In ...
A weak acid cannot always be neutralized by a weak base, and vice versa. However, for the neutralization of benzoic acid (K a,A = 6.5 × 10 −5) with ammonia (K a,B = 5.6 × 10 −10 for ammonium), K = 1.2 × 10 5 >> 1, and more than 99% of the benzoic acid is converted to benzoate.
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 ...
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 ...
The Avogadro constant, commonly denoted N A [1] or L, [2] is an SI defining constant with an exact value of 6.022 140 76 × 10 23 mol −1 (reciprocal moles). [3] [4] It is this defined number of constituent particles (usually molecules, atoms, ions, or ion pairs—in general, entities) per mole and used as a normalization factor in relating the amount of substance, n(X), in a sample of a ...
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
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: