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In aqueous solution, ammonia deprotonates a small fraction of the water to give ammonium and hydroxide according to the following equilibrium: . NH 3 + H 2 O ⇌ NH + 4 + OH −.. In a 1 M ammonia solution, about 0.42% of the ammonia is converted to ammonium, equivalent to pH = 11.63 because [NH +
Heating at higher temperatures results in decomposition into ammonia, nitrogen, sulfur dioxide, and water. [17] As a salt of a strong acid (H 2 SO 4) and weak base (NH 3), its solution is acidic; the pH of 0.1 M solution is 5.5. In aqueous solution the reactions are those of NH + 4 and SO 2−
The relative activity of a species i, denoted a i, is defined [4] [5] as: = where μ i is the (molar) chemical potential of the species i under the conditions of interest, μ o i is the (molar) chemical potential of that species under some defined set of standard conditions, R is the gas constant, T is the thermodynamic temperature and e is the exponential constant.
For instance, a 5 × 10 −8 M solution of HCl would be expected to have a pH of 7.3 based on the above procedure, which is incorrect as it is acidic and should have a pH of less than 7. In such cases, the system can be treated as a mixture of the acid or base and water, which is an amphoteric substance.
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]
Ammonia is moderately basic; a 1.0 M aqueous solution has a pH of 11.6, and if a strong acid is added to such a solution until the solution is neutral (pH = 7), 99.4% of the ammonia molecules are protonated. Temperature and salinity also affect the proportion of ammonium [NH 4] +.
The following table lists the Van der Waals constants (from the Van der Waals equation) for a number of common gases and volatile liquids. [ 1 ] To convert from L 2 b a r / m o l 2 {\displaystyle \mathrm {L^{2}bar/mol^{2}} } to L 2 k P a / m o l 2 {\displaystyle \mathrm {L^{2}kPa/mol^{2}} } , multiply by 100.
The solubility of a specific solute in a specific solvent is generally expressed as the concentration of a saturated solution of the two. [1] Any of the several ways of expressing concentration of solutions can be used, such as the mass, volume, or amount in moles of the solute for a specific mass, volume, or mole amount of the solvent or of the solution.