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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 +
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]
In a solution of potassium hydrogen iodate KH(IO 3) 2 at 0.02 M the activity is 40% lower than the calculated hydrogen ion concentration, resulting in a much higher pH than expected. When a 0.1 M hydrochloric acid solution containing methyl green indicator is added to a 5 M solution of magnesium chloride, the color of the indicator changes from ...
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.
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 standard hydrogen electrode (SHE), with [ H +] = 1 M works thus at a pH = 0. At pH = 7, when [ H +] = 10 −7 M, the reduction potential of H + differs from zero because it depends on pH. Solving the Nernst equation for the half-reaction of reduction of two protons into hydrogen gas gives: 2 H + + 2 e − ⇌ H 2
pH = 1 / 2 pK w + 1 / 2 log (1 + T A / K a ) With a dilute solution of the weak acid, the term 1 + T A / K a is equal to T A / K a to a good approximation. If pK w = 14, pH = 7 + (pK a + log T A)/2. This equation explains the following facts: The pH at the end-point depends mainly on the strength of the ...
The same direct relationship applies to gases and vapors diluted in air for example. Although, thorough mixing of gases and vapors may not be as easily accomplished. [citation needed] For example, if there are 10 grams of salt (the solute) dissolved in 1 litre of water (the solvent), this solution has a certain salt concentration . If one adds ...