By Heike Kahlert, Fritz Scholz

Understanding acid-base equilibria made effortless for college kids in chemistry, biochemistry, biology, environmental and earth sciences. fixing chemical difficulties, be it in schooling or in actual lifestyles, frequently calls for the certainty of the acid-base equilibria in the back of them. in response to decades of training adventure, Heike Kahlert and Fritz Scholz current a strong device to satisfy such demanding situations. they supply an easy advisor to the basics and purposes of acid-base diagrams, heading off advanced arithmetic. This textbook is richly illustrated and has complete colour all through. It bargains studying good points akin to boxed effects and a set of formulae.

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**Extra resources for Acid-Base Diagrams**

**Example text**

Fig. 45), the following charge balance must hold: cH3 Oþ ¼ cOHÀ À cHB (62) This means that the pH of the solution must be situated slightly to the right of the pH-coordinate of P1 in Fig. 45, since on the left side of P1 the sum cH3 Oþ þ cHB is always larger than the concentration OHÀ (in other words, the overall concentration of OHÀ results from the protolysis of BÀand autoprotolysis of water). The concentration of H3 Oþ at P1 is about 10À13 mol LÀ1, and so it can be neglected in the charge balance, and the pH is given by the pH-coordinate of the crossing point of the OHÀ and ethanol (HB) lines: cOHÀ % cHB (78) The diagram also reveals that the equilibrium concentration of ethanolate ions can be neglected in relation to the ethanol concentration, so that the amount balance can be approximated by the relationship: cHB % C BÀ (106) This makes it possible to derive a simplified equation by using the Eqs.

1 mol L Fig. 14 Plotting the two asymptotes of BÀ with Eq. (35) for pH < pKa and Eq. (36) for À1 pH > pKa . 2 Dibasic Acids In the case of dibasic acids H2 B; the following two protolysis reactions have to be considered: H2 B þ H2 O ! HBÀ þ H3 Oþ (37) HBÀ þ H2 O ! B2À þ H3 Oþ (38) The equilibrium equation (37) is characterized by the following law of mass action: Ka1 ¼ cHBÀ cH3 Oþ cH2 B (39) cB2À cH3 Oþ cHBÀ (40) and equilibrium equation (38) by: Ka2 ¼ As an analogy to Eq. (18), the following amount balance has to be formulated: À C H2 B ¼ cH2 B þ cHB þ cB2À (41) The mathematical equations of the functions are given in the Appendix.

B4À þ H3 Oþ (48) The mathematical equations of the functions are given in the Appendix. The plotting of all species in the diagram can be made with the help of the line equations given in the Appendix. Following the procedures for the mono- to tribasic acids, this is stepwise made in Figs. 26, 27, 28, 29, 30, and 31. The line of H4 B has the slope À2 in the range pKa2 < pH < pKa3 (H2 B2À is dominating). The slope is À3 in the range pKa3 < pH < pKa4 (HB3À is dominating). Sometimes, also in the example shown in Fig.