Digestion of food involves breaking down of good componets into smaller molecules by enzymes. These enzymes are active only at certain hydrogen ion concentration. As a result, certain food combinations can facilitate or retard the process of digestion. Of the following combinations, one that can result in very efficient digestion is

Consider a solution of CH_3COONH_4 which is a salt weak acid and weak base. The equilibrium involved in the solutions are : CH_3COO^(-)+H_2OhArrCH_3COOH+OH^(-) " ........"(i) NH_(4)^(+)+H_(2)hArrNH_(4)OH+H^(+)" ........"(ii) H^(+)+OH^(-)hArrH_(2)O" ........"(iii) If we add these reactions, then the net reaction is : CH_(3)COO^(-)+H_(2)^(+)+H_(2)OhArrCH_(3)COOH+NH_(4)OH" ........"(iv) Both CH_(3)COO^(-) and NH_(4)^(+) get hydrolysed independently and their hydrolysis depends on : (a) their initial concentration (b) The value of K_(h) which is (K_(w))/(K_(a)) for CH_(3)COO^(-) and (K_(w))/(K_(b)) for NH_(4)^(+) . Since both of the ions were produced form the salt, their initial concertration are same. Therefore, unless and until the value of (K_(w))/(K_(a)) or K_(a) and K_(b) is same, the degree of hydrogen of ions can't be same. To explain why we assume that degree of hydrolysis of cation and anion is same, we needed to now look at the third reaction i.e.,combination of H^(+) and OH^(-) ions. It is obvious that this reaction happens only because one reaction produced H^(+) ion and the other prodcued OH^(-) ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this in mind that the equilibrium which has smaller value of the equilibrium constant is affected more by the common ion effect. For the same reason if for any reson a reaction is made to occur to a greater extent by the consumption of any of the prodcut ion, the reaction with the smaller value of equilibrium constant tends to get affected more. Therefore, we conclude that firstly the hydroylsis of both the ions occurs more in the presence of each other (due to consumption of the product ions) than in each other's absence. Secondly,the hydroylsis of the ion which occurs to a lesser extent (due to smaller value of K_(h)) is affected more than the one whole K_(h) is greater. Hence we can see that the degree of hydroylsis of both the ions would be close to each other when they are getting hyderolysed in the presence of each other. In the hydrolysis of salt weak acid and weak base :

Consider a solution of CH_3COONH_4 which is a salt weak acid and weak base. The equilibrium involved in the solutions are : CH_3COO^(-)+H_2OhArrCH_3COOH+OH^(-) " ........"(i) NH_(4)^(+)+H_(2)hArrNH_(4)OH+H^(+)" ........"(ii) H^(+)+OH^(-)hArrH_(2)O" ........"(iii) If we add these reactions, then the net reaction is : CH_(3)COO^(-)+H_(2)^(+)+H_(2)OhArrCH_(3)COOH+NH_(4)OH" ........"(iv) Both CH_(3)COO^(-) and NH_(4)^(+) get hydrolysed independently and their hydrolysis depends on : (a) their initial concentration (b) The value of K_(h) which is (K_(w))/(K_(a)) for CH_(3)COO^(-) and (K_(w))/(K_(b)) for NH_(4)^(+) . Since both of the ions were produced form the salt, their initial concertration are same. Therefore, unless and until the value of (K_(w))/(K_(a)) or K_(a) and K_(b) is same, the degree of hydrogen of ions can't be same. To explain why we assume that degree of hydrolysis of cation and anion is same, we needed to now look at the third reaction i.e.,combination of H^(+) and OH^(-) ions. It is obvious that this reaction happens only because one reaction produced H^(+) ion and the other prodcued OH^(-) ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this in mind that the equilibrium which has smaller value of the equilibrium constant is affected more by the common ion effect. For the same reason if for any reson a reaction is made to occur to a greater extent by the consumption of any of the prodcut ion, the reaction with the smaller value of equilibrium constant tends to get affected more. Therefore, we conclude that firstly the hydroylsis of both the ions occurs more in the presence of each other (due to consumption of the product ions) than in each other's absence. Secondly,the hydroylsis of the ion which occurs to a lesser extent (due to smaller value of K_(h)) is affected more than the one whole K_(h) is greater. Hence we can see that the degree of hydroylsis of both the ions would be close to each other when they are getting hyderolysed in the presence of each other. For 0.1 M CH_3COONH_(4) salt solution given, K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=2xx10^(-5) . In the case : degree of hydrolysis of cation and anion are :

Strong acid versus strong base: The principle of conductometric titrations is based on the fact that during the titration, one of the ions is replaced by the other and invariable these two ions differ in the ionic conductivity with the result that thhe conductivity of the solution varies during the course of the titration. take, for example, the titration between a strong acid, say HCl, and a string base, say NaOH before NaOH is added, the conductance of HCl solution has a high value due to the presence of highly mobile hydrogen ions. As NaOH is added, H^(+) ions are replaced by relatively slower moving Na^(+) ions. consequently the conductance of the solution decreases and this continues right upto the equivalence point where the solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution contains a excess of the fast moving OH^(-) ions with the result that its conductance is increased ad it condinues to increase as more and more of NaOH is added. If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown in Fig. The descending portion AB represents the conductances before the equivalence point (solution contains a mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which represent the minium conductance is due to the solution containing only NaCl with no free acid or alkali and thus represents the equivalence point. this point can, however, be obtained by the extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point expermentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators. Weak acid versus strong base: Let us take specific example of acetic acid being titrated against NaOH . Before the addition of alkali, the solution shows poor conductance due to feeble ionization of acetic acid. Initially the addition of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the dissociation of acetic acid due to the common ion Ac^(-) and thus the conductance of the solution decreases in the beginning. but very soon the conductance start increasing as addition of NaOH neutralizes the undissociated HAc to Na^(+)Ac^(-) thus causing the replacement of non-conducting HAc with Strong-conducting electrolyte Na^(+)Ac^(-) . the increase in conductance continuous right up to the equivalence point. Beyond this point conductance increases more rapidly with the addition of NaOH due to the highly conducting OH^(-) ions, the graph near the equivalence point is curved due to the hydrolysis of the salt NaAc . The actual equivalence point can, as usual, be obtained by the extrapolation method. In all these graphs it has been assumed that the volume change due addition of solution from burrette is negnigible, hence volume change of the solution in beaker the conductance of which is measured is almost constant throughout the measurement. Q. The nature of curve obtained for the titration between weak acid versus strong base as described in the above passage will be: