# dissociation constant in electrochemistry

Salts such as $$K_2O$$, $$NaOCH_3$$ (sodium methoxide), and $$NaNH_2$$ (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table $$\PageIndex{2}$$, are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of $$OH^−$$ and the corresponding cation: $K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \label{16.5.18}$, $NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \label{16.5.19}$, $NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \label{16.5.20}$. Conversion of chemical energy into electrical energy. Thus the numerical values of K and $$K_a$$ differ by the concentration of water (55.3 M). Butyric acid is responsible for the foul smell of rancid butter. For the general electrochemical reaction of the type: Ecell = Eocell – RT/ nF ln [C]c [D]d / [A]a [B]b, Ecell = Eocell – 2.0303 RT / nF log [C]c [D]d / [A]a [B]b, Ecell = Eocell – 0.0591 / n log [C]c [D]d / [A]a [B]b at 298K. On a vérifié l'accord entre les vagues théoriques et celles expérimentales dans l'acétonitrile et dans des solutions aqueuses de HClO4. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. Cell potential or emf of a cell: In the galvanic cell there are two half cell, the oxidation half-cell( anode) and the reduction half -cell( cathode).Due to the difference in the potentials of these half-cells, the electric current moves from the electrode of higher potential (cathode) to the electrode of lower potential( anode). If Eocell is negative then the reaction is not feasible. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, in the electrolysis of molten sodium chloride, sodium chloride is melted (above 801oC), two electrodes are inserted into the melt, and an electric current is passed through the molten salt. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. In particular, we would expect the $$pK_a$$ of propionic acid to be similar in magnitude to the $$pK_a$$ of acetic acid. E.g. Electrochemical cell are of two types: galvanic cells and electrolytic cells. Copyright © 1967 Published by Elsevier Ltd. https://doi.org/10.1016/0013-4686(67)80104-X. Conversely, smaller values of $$pK_b$$ correspond to larger base ionization constants and hence stronger bases. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base. When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte. The standard conditions taken are: A series of the standard electrode has been established by measuring the potential of various electrodes versus standard hydrogen electrode(SHE). When the potential difference is applied between the two electrodes, Na+ and H+ ions move towards the cathode and Cl– and OH– ions move towards the anode. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths. The $$pK_a$$ and $$pK_b$$ for an acid and its conjugate base are related as shown in Equations $$\ref{16.5.15}$$ and $$\ref{16.5.16}$$. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair: $\underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}}$. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair: $\underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}}$. The conjugate base of a strong acid is a weak base and vice versa. Because the $$pK_a$$ value cited is for a temperature of 25°C, we can use Equation $$\ref{16.5.16}$$: $$pK_a$$ + $$pK_b$$ = pKw = 14.00. i.e., the generation of electricity by spontaneous redox reactions. Smaller values of $$pK_a$$ correspond to larger acid ionization constants and hence stronger acids. The values of $$K_a$$ for a number of common acids are given in Table $$\PageIndex{1}$$. At the bottom left of Figure $$\PageIndex{2}$$ are the common strong acids; at the top right are the most common strong bases. From Table $$\PageIndex{1}$$, we see that the $$pK_a$$ of $$HSO_4^−$$ is 1.99. Calculate $$K_a$$ and $$pK_a$$ of the dimethylammonium ion ($$(CH_3)_2NH_2^+$$). (In fact, the $$pK_a$$ of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) The oxidation and reduction take place in two separate compartments. You will notice in Table $$\PageIndex{1}$$ that acids like $$H_2SO_4$$ and $$HNO_3$$ lie above the hydronium ion, meaning that they have $$pK_a$$ values less than zero and are stronger acids than the $$H_3O^+$$ ion. For example, in the electrolysis of NaCl solution, apart from Na+ and Cl–ions the solution of sodium chloride also contains H+ and OH– ions due to ionisation of water. The equilibrium constant for this dissociation is as follows: $K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \label{16.5.2}$. In contrast, in the second reaction, appreciable quantities of both $$HSO_4^−$$ and $$SO_4^{2−}$$ are present at equilibrium. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Then refer to Tables $$\PageIndex{1}$$and$$\PageIndex{2}$$ and Figure $$\PageIndex{2}$$ to determine which is the stronger acid and base. Legal. Conversely, the conjugate bases of these strong acids are weaker bases than water. As you learned, polyprotic acids such as $$H_2SO_4$$, $$H_3PO_4$$, and $$H_2CO_3$$ contain more than one ionizable proton, and the protons are lost in a stepwise manner. Calculate $$K_a$$ for lactic acid and $$pK_b$$ and $$K_b$$ for the lactate ion. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. If we add Equations $$\ref{16.5.6}$$ and $$\ref{16.5.7}$$, we obtain the following: In this case, the sum of the reactions described by $$K_a$$ and $$K_b$$ is the equation for the autoionization of water, and the product of the two equilibrium constants is $$K_w$$: Thus if we know either $$K_a$$ for an acid or $$K_b$$ for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair. 5. For example, nitrous acid ($$HNO_2$$), with a $$pK_a$$ of 3.25, is about a million times stronger acid than hydrocyanic acid (HCN), with a $$pK_a$$ of 9.21. 1 molar concentration of each ion in the solution. Electrode potential: In a galvanic cell, when two-electrode are dipped in their respective ion there is a tendency of one of the electrodes (anode) to undergo oxidation whereas the ion of the other electrode (cathode) has the tendency to gain an electron. What is the electrochemical series? It is used to maintain the charge balance and to complete the circuit by allowing the flow of ions through it. Identify the conjugate acid–base pairs in each reaction. The base ionization constant $$K_b$$ of dimethylamine ($$(CH_3)_2NH$$) is $$5.4 \times 10^{−4}$$ at 25°C. Equilibrium always favors the formation of the weaker acid–base pair. Measurements of the conductivity of 0.1 M solutions of both HI and $$HNO_3$$ in acetic acid show that HI is completely dissociated, but $$HNO_3$$ is only partially dissociated and behaves like a weak acid in this solvent. Consider, for example, the $$HSO_4^−/ SO_4^{2−}$$ conjugate acid–base pair. Calculate $$K_b$$ and $$pK_b$$ of the butyrate ion ($$CH_3CH_2CH_2CO_2^−$$). Typically, electrochemistry deals with the overall reactions when multiple redox reactions occur simultaneously, connected via some external electric current and a suitable electrolyte. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. On supplying current the ions move towards electrodes of opposite polarity and simultaneous reduction and oxidation take place. These interconversions are carried out in equipment called electrochemical cell. In fact, all six of the common strong acids that we first encountered in Chapter 4 have $$pK_a$$ values less than zero, which means that they have a greater tendency to lose a proton than does the $$H_3O^+$$ ion.

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