How Do Acid-Base Equilibria Work?

AI Thread Summary
The discussion centers on understanding acid-base equilibrium equations and the associated terminology. Key points include the identification of variables: A represents an acid, B a base, and H denotes hydrogen ions. The notation HA refers to a monoprotic acid, while BH+ indicates a basic salt, such as the ammonium ion. The equations illustrate the reversible reactions where acids donate protons (H+) to water, forming hydronium ions (H3O+) and anions (A-). The equilibrium constants (K and Ka) are derived from these reactions, with the concentration of water often neglected due to its high concentration. The conversation emphasizes the importance of grasping these fundamental concepts to fully understand acid-base equilibria, noting that the equilibrium constants span a wide range, making the use of their negative logarithms more practical for analysis.
Huzaifa
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Equilibrium constants in acid-base reactions equations
$$
\begin{array}{c}
\mathrm{HA}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \\
\mathrm{K}=\dfrac{\left[\mathrm{A}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left.[\mathrm{HA}] \mathrm{H}_{2} \mathrm{O}\right]} \quad \mathrm{K}_{\mathrm{a}}=\dfrac{\left[\mathrm{A}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{[\mathrm{HA}]} \\
\mathrm{BH}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{B}+\mathrm{H}_{3} \mathrm{O}^{+} \\
\mathrm{K}=\dfrac{[\mathrm{B}]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{BH}^{+}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]} \quad \mathrm{K}_{\mathrm{a}}=\dfrac{[\mathrm{B}]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{BH}^{+}\right]} \\
\mathrm{HA}+\mathrm{B} \rightleftharpoons \mathrm{A}^{-}+\mathrm{BH}^{+} \\
\mathrm{K}_{\mathrm{c}}=\dfrac{\left[\mathrm{A}^{-}\right]\left[\mathrm{BH}^{+}\right]}{[\mathrm{HA}][\mathrm{B}]}
\end{array}
$$

Hello! I am not able to understand these equations for acid base equilibrium please. I have attached the images in the following post.
 
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What you posted is a definition - nothing to understand here, it just is what it is, all you have to do is to accept it - and some algebraic rearranging, that is not difficult to follow if you have a basic understanding of HS math. You have not explained what it is that you don't understand.
 
Borek said:
What you posted is a definition - nothing to understand here, it just is what it is, all you have to do is to accept it - and some algebraic rearranging, that is not difficult to follow if you have a basic understanding of HS math. You have not explained what it is that you don't understand.
I don't understand A, B, and H, and the positive and negative signs in superscripts. I assume that A is Acid, and B is Base, and H is Hydrogen. Please correct me if I am wrong.

I also don't understand HA and BH+? Also there is no OH-.
 
Huzaifa said:
the positive and negative signs in superscripts
These are ion charges. I suggest you you get back to basics, otherwise you won't be able understand the acid/base equilibrium.

A and B are often used as general symbols for acid (base) residue. HA means just any monoprotic acid.
 
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The first equation shows the reversible transfer of the H+ ion from the acid HA to water H2O to form the hydronium ion H3O+ and the anion A-. The second shows the calculation of the equilibrium constant K for this reaction. Since the concentration of water is very high (~55 molar) it does not change significantly and is neglected in the calculation of the acid equilibrium constant Ka. Thus the equilibrium constant is equal to the concentration of the anion times the concentration of the hydronium ion divided by the concentration of the acid and of the water.
The second set of equations refers to the reaction of a basic salt such as the ammonium ion NH4+ with water and the third set refers to the reaction of an acid HA and a base such as ammonia NH3. These equilibrium constants cover 14 powers of ten from 101 to 10-14 so it is more convenient to use their negative logarithms: 0 to 14.
 
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