Acid salts are a class of salts that produce an acidic solution after being dissolved in a solvent. Its formation as a substance has a greater electrical conductivity than that of the pure solvent. An acidic solution formed by acid salt is made during partial neutralization of diprotic or . A half-neutralization occurs due to the remaining of replaceable hydrogen from the partial dissociation of weak acids that have not been reacted with () to create water molecules.
For example, the acid salt ammonium chloride is the main species formed upon the half neutralization of ammonia in aqueous solution of hydrogen chloride:
+Example of acid salts |
An acid salt can be mixed with certain base salt (such as sodium bicarbonate or baking soda) to create baking powders which release carbon dioxide. Leavening agents can be slow-acting (e.g. sodium aluminum phosphate) which react when heated, or fast-acting (e.g., cream of tartar) which react immediately at low temperatures. Double-acting baking powders contain both slow- and fast-acting leavening agents and react at low and high temperatures to provide leavening rising throughout the baking process.
Disodium phosphate, , is used in foods and monosodium phosphate, , is used in animal feed, toothpaste and evaporated milk.
Other possible factors that could vary pH level of a solution are the relevant equilibrium constants and the additional amounts of any base or acid.
For example, in ammonium chloride solution, is the main influence for acidic solution. It has greater value compared to that of water molecules; of is , and of is . This ensures its deprotonation when reacting with water, and is responsible for the pH below 7 at room temperature. will have no affinity for nor tendency to hydrolyze, as its value is very low ( of is ).Lower, S.K., (1999). Introduction to acid-base chemistry. Chem1 General Chemistry Text. Retrieved from http://www.chem1.com/acad/pdf/c1xacid1.pdf
Hydrolysis of ammonium at room temperature produces:
K_\mathrm{a} = \frac{\mathrm{[NH_3][H_3O^+]}}{\mathrm{[NH_4^+]}} = \frac{K_\mathrm{w}}{K_\mathrm{b}}
K_\mathrm{a} = \frac{[1.0 \times 10^{-14}]}{[1.8 \times 10^{-5}]} = {5.6} \times 10^{-10}
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