Chapter 5. Acid/Base Reactions

Transcription

1 hapter 5. cid/base Reactions 5. Log p Diagrams (Graphical Solutions) 6. itration 7. Buffers and Buffer Intensity 8. arbonate System 7 Water hemistry 5. Log p Diagrams p as an independent variable Develop equations for the concentration of a given species as a function of p and known quantities (e.g., w,, ) a. Introduction to log -p diagram : weak monoprotic acid example i. Draw + line -log + = p log + = -p ii. Draw O - line + O - = w log + + log O - = log w log O - = p p w

2 iii. Draw c and c - lines c = c- = If p << p or + >> log c = log log c - = log p + p If p >> p or + << log c = log + p p log c- = If p = p, + = log c = log log = log. log c- = log.. M c Example - =.8* -5 ; p = log c = -.7;.5 p - log c- = p; -.7 BE or PE : + = O - + c- Look at intersections -p 7 - p.; c >> O-, + = c - System point: where p = p & log = log

3 b. Rapid method for construction of log-p diagram i. Plot diagonal + and O - lines. ii. Draw a horizontal line corresponding to log. iii. Locate system point. -p = p & log = log - make a mark. units below system point iv. Draw 45 lines (slope ) below line, and aimed at system point. v. pproximate curved sections of species lines p unit around system point. vi. Repeat steps as necessary for more complex graphs. #-5 for additional p s of polyprotic acids. #-5 for other acid/base pairs. c. Log-p diagram for a monoprotic weak base N + O = N 4+ + O O - = O N + + = N the same as drawing a diagram for conjugate acid - p for the system point = p = p W p B = = M N

4 4 Example 4. :. M N 4 l - an add log l - - +, O -, N, N 4+, l - BE : + + N 4+ = O - + l - PE : + = O - + N ; Usually easier When + = N, p = 5.8 hecking BE: N 4+ = l - ; +, O - negligible Example 4. :. M Potassium acetate (c) - an add log + - +, O -, c, c -, + BE: = O - + c - PE: + + c = O - -Point when + = O -, not good - Good when c = O -, p of BE satisfied. d. Log-p diagram for a weak acid and weak base - Mixtures of. M acetic acid (c) and.m ammonium hydroxide (N 4 O) - hey may exist in relatively high concentration in anaerobic digesters - Superimpose the curves for each material

5 5 Example 4.4 N 4 l + c ase. M each -Species: +, O -, c, c -, N 4+, l - -BE: + + N 4+ = O - + c - + l - -PE: + PRL O c N 4 O - c- N + = O - + c - + N + = O- does not work + = c- O. p =.8 6. itration a. itration of strong acids and bases itration: procedure by which a measured amount of chemical or reagent is added to a solution in order to bring about a desired and measured change. - When strong base is titrated, + p changes very little at first, then slowly declines to a p of about. + essentially vertical between p values of and 4. + equivalence point or stoichiometric end point lies between these values. - Equivalence point is reached when the equivalents of base (or acid) added equal the equivalents of acid (or base) initially present.

6 6 Example: Strong acid l is titrated with strong base NaO NaO + l O + Na + + l - BE: Na = O - + l - = conc. of l V = initial volume of l = conc. of NaO V = vol. of NaO added V V O V V V V Moving terms related to titration to the left, V V ( ) O V V V V w ( f ) V V f = equivalence fraction. t the equivalence point, f =. Example ml of. l is titrated with.5 M NaO alculate p values at different points in the titration. V V ( ) O V V V - V negligible. 5 times concentrated. Very small addition. -O - negligible at lower p region V V ( ) V V

7 7 Example 4.8. ontinued alculate p values at different points in the titration. V V ( ) O V V V - V negligible. 5times concentrated. Very small addition. -O - negligible at lower p region 5 ml addition: 5 % neutralized. Use V/ V. p. 9 ml: 9% neutralized. p. ml: % neutralized. Equivalence point! ave only Nal and water. Nal is neutral, p is 7. fter the equivalence point, V V ( ) V V V V W ( ) V V V O ml: p = When p = 4, 99% neutralized. Practically the titration is complete. b. itration of weak acids and bases - Frequently encountered in natural water systems. - When weak acids are titrated with strong bases, the character of the titration curve depends on whether the acid is monoprotic or polyprotic. - Look at different (p ) values. Shapes (buffering well at p ) - PO 4 example - multiprotic

8 8 V V V itration of a weak monoprotic acid with a strong base = O - = w ) ) + B + + O - B O (B + : cation associated with the strong base) BE : + + B + = - + O - ) - Simultaneous solution of ), ), ) can determine + -Introduce, V,, V - MB on weak acid: + - = V =, : it changes as you add base. V V -MB on B + V B + = V V - Introduce ionization fraction ( V V ) ( ) V ( V V ) ( ) V B+ -BE becomes V V O ( V V V V ) - = *, - Derive equation for f by multiplying (V o +V)/ o V o V V V f ( )( w ) V V Example 4. 9

9 9 c. Estimation of p during titration ase: Weak acid. = o. itration with a strong base i. Beginning of titration + B + + O - B O BE: + + B + = - + O- + + B + = - + O equation - << ( ) p log log log ( p log ) Example heck the + - assumption from the p value obtained. - It assumption is not correct, use the approximation ( ) ii. Midpoint of titration + B + + O - B O - - increases and decreases - When 5% completed, B + ; diluted slightly onc. of B + & - become significant + << B +, O - << - BE: + + B + = - + O - -BE becomes B + - ) ) - ) & ) to mass action equation ( ) p p ; indicates the value of expressing the ionization constant in the p form.

10 iii. Equivalence point of titration -B + = o, + << O- ; + B + + O - B O BE: + + B + = - + O - - B O - O but from MB hus O - = W W - = o W p (log p p w ) - Depends on concentration of acid and equilibrium constant. - he weaker the acid and the higher its conc., the higher the equivalence point p. p of weak base during titration with strong acid - an be determined in a similar fashion. - Beginning of titration: p p w p B log - Midpoint of titration: p p w p B - Equivalence point of titration: p ( p w p log B ) - he inflection point at the equivalence point in the titration of weak acids or bases with ionization constants less than about -7 M or p values greater than 7 is not sharp and so is difficult to detect accurately.

11 itration curves for weak bases -Weak bases - Salts of weak acids are alkaline, Nac, Na O - Simply acid titration curves in reverse; curve is a mirror image of that for the acid. Example 4. - p of the equivalence point for each ionization is approximately equal to the average of the p value for that ionization and the p value for the following ionization. Example 4. d. Use of log -p diagram for titrations - cetic acid example:. M M c (c + B + + O - B + + c - + O) BE: + + B + = c - + O- - Beginning of titration, + c - - Midpoint, B + c - = c => located at p ; p regardless of o - Endpoint, B + = o = c + c - (to BE) + + c + c - = c - + O c = O - - Effect of changes in the conc. of weak acid: observed by moving the horizontal c and c - curves up or down. - Study more examples in textbook.

12 7. Buffers and Buffer Intensity Buffers: Substances in solution that offer resistance to changes in p (as acids or bases are added to or formed within the solution) - ommonly a mixture of an acid and its conjugate base. Best to have a reservoir of each so there is a resistance to change in both direction ( for base, - for acid). - Poorly buffered waters are susceptible to acid precipitation. Buffer intensity: the amount of strong acid or base required to cause a small shift in p. - Midpoint in the titration curve => slope is at a minimum => intensity greatest. - t the midpoint, the solution contains equal quantities of the ionized and the unionized acid or base. - p at the midpoint in the titration => p or p W -p B for weak bases. hus use able 4. & 4.; - Buffers effective at a p near the p value; but effective within p unit of p enderson-asselbalch equation (Equilibrium for a weak acid): p p salt log acid - p of a buffer solution made of a weak acid and its salt depends upon the ratio of salt concentration to acid concentration. - Solution to weak acid/conjugate base problem using equil., MBE, and BE gives the same answer. -p for phosphoric acid ( PO - 4 = + + PO - 4 ) is 7. Useful buffer at neutral p. ommonly used in analytical tests.

13 Example p of a buffer containing.m acetic acid &. M sodium acetate salt p = p p log = p + log (./.) = = 4.74 acid. ap after l is added to this solution to give a concentration of. M. Salt neutralizes l to produce acid. p = log Buffer intensity (index) calculation db d ; number of moles of acid or base required to produce a given dp dp change in p; highest at the midpoint; low at the endpoint Buffer Index - c example - an be readily determined from a titration curve. lso can be calculated. - moles of monoprotic acid & B moles of base: c + Nac BE: B + + = O Use ionization fraction expression for - lso, Since w B db db d dp d dp ln p log. d ln d dln d dp ;... d dp db db. dp d - Differentiating B with respect to + and substituting into the above eq. yields, W. ( )

14 4 Example Buffer containing. M c and. M acetate. p = 5. - mount of NaO (mol/l) required to increase the p to 5.? =.+.=.M + = -5 Use W. ( ) =.6 mol/l of base per p unit NaO required = dp = (.6)(5.-5.) =.6 mol/l 8. arbonate System a. Significance - Regulating p in natural waters (major acid-conjugate base systems) - Sources: volcanism, combustion, respiration. - Sinks: photosynthesis, autotropic metabolism, precipitation - Major forms of carbon on earth

15 5 Mineral arbonation of arbon Dioxide Mineral carbonation refers to the conversion of silicates to solid carbonates, mimicking the natural process by which O is removed from the atmosphere O (g) + asio weathering O - (aq) + a + (aq) + SiO mineralisation ao (s) Primary process by which carbon dioxide is removed from the atmosphere >99% world s carbon reservoir is locked up as limestone & dolomite rock ao & MgO hermodynamically favourable, but kinetically slow ~ tonnes O in atmosphere ~ billion tonnes/year O ~ 8 tonnes O in carbonate rock OMN: 7,km of % olivine; sufficient to mineralise centuries of global O emissions. ombine Mineral arbonation with Wastes reatment

16 6 b. arbonate species and their equilibria - Gas transfer and acid-base reactions p = 6.7, p =. O (g) = O (aq) = O (aq)/p co = = -.5 (M/atm) = enry s law constant enry s law (p.5): Gas transfer (gas-liquid) equilibrium. he mass of any gas that will dissolve in a given volume of a liquid, at constant temperature, is directly proportional to the pressure that the gas exerts above the liquid. P gas equil his form more common (careful about the unit). =.6 (atm/m) b. arbonate species and their equilibria - continued O (aq) + O = O O = = -.8 (hydration equil.) O = + + O - O = O O ( aq ) O.5 O - It is difficult to distinguish between O and O by analytical procedure => ypothetical species introduced O * = O +O (aq) O (aq) ( O (aq) = 6 O : total analytical conc. is essentially dissolved carbon dioxide) = O O O 6.7 * O O O O( aq ) O = O. O

17 7 c. alculation of carbonate species in open and closed systems i. losed system: log-p diagram (for polyprotic acids/bases). M O Example = O * + O - + O - = * O O. O O * O O O - Diagram is constructed just as for monoprotic acids and bases, except that the slope of line for O - changes from + to + when the p drops below p. + he last term in the carbonate eq. becomes dominant when + >> - ow do you determine equilibrium p? Use proton condition: + = O - + O - + O - ; same as charge balance * O O O p p p O log log

18 8 ii. Open system : Log - p diagram for carbonate species in equilibrium with the O gas (.6 atm) in the atmosphere - Other examples: N, S O * O (aq) = P co / = -.5 /.6 = -5 M; independent of p From the first dissociation eq., 5 O - = log O - = 5 p = p -.7 Similarly, log O - = p.7 = O * + O - + O - Predominance + When p < p, O * predominant + When p < p < p, O - predominant + When p > p, O - predominant - increases rapidly as p increases above p Open system Examples i. Deionized water BE or PE (ssuming O * as an initial species) + = O - + O - + O - p of 5.68 (lean rain in equilibrium with O ) Exact Solution in Example 4.6 ii. -5 M NaO Let s do charge balance. + + Na+ = O - +O - + O - Na+ = -5 = O - omework #4 4.8, 4.4, 4.4, Determine p and conc. of carbonate species in water containing -5 M Na O in an open system. PO = -.5 atm

19 9 d. lkalinity and acidity (p ) i. lkalinity: apacity of water to neutralize acids - form of acid neutralizing capacity (N: Example 4.7) - itration curve for a hydroxide-carbonate mixture - Interpretation in most natural waters: lk tot = O - + O - + O Net deficiency of protons with respect to O ( O *): proton condition concept - In sea water we use: lk tot = O - + O - + B(O) 4- + PO SiO 4- + MgO - +O Measurements of lkalinity - Measurement by titration with a strong acid (back) to p about 4.5. ) Phenolphthalein endpoint: arbonate alkalinity + + O - O + + O - O - light pink to colorless at p about 8. ) Methyl orange endpoint: otal alkalinity + + O - O Yellow to red at p about 4.5, where all carbonate are as O - Unit: meq/l or mg/l as ao ii. cidity - apacity of water to neutralize bases - form of base neutralizing capacity (BN) - Interpretation in most natural waters: cytot = O + O O - : Net excess of protons with respect to O - - Measurement by titration with a strong base back to the p of a pure O - solution (about.7)

20 Example - ml of a water required. ml of. N SO 4 to titrate to the phenolphthalein endpoint -.9 ml of. N SO 4 to titrate it further to the methyl orange endpoint. - otal and carbonate alkalinity? arbonate alkalinity =.eq meq meq (.ml)( )( )( ). L q ml L.eq meq meq otal alkalinity = ( 4mL)( )( )( ). 4 as ao? L q ml L Example groundwater sample that was carbonated with minerals such as ao (s), MgO (s), etc. - Sampled without exposing the water to atmosphere. No other weak acids or bases. - nalysis shows that,o is - M and p = 7.6 a) cid Neutralizing apacity? lkalinity N = lkalinity = O - + O - + O - + ( O - & O - can be determined if you know and p) = ( + ) = - (.94+(.76E-)) = 9.47* -4 eq/l = 47 mg/l as ao b) p when GW is brought into equilibrium with the atmosphere (P co = -.5 atm) - Use BE to develop the master equation. B + + = O - + O - + O - + B = sum of the strong-base cations (Na +, a +, etc.) = sum of the strong-acid anions (l -, NO - ) B - : Initially present in the water and not volatile; Does not change with exposure to O. B - = O - + O - + O - + = 9.47* -4 = alkalinity lkalinity does not change upon O exposure. 9.47* -4 = O- + O - + O - + o solve for +, obtain expressions for carbonate species that can take into account of the open system. changes (No fixed value for now) according to p. O * = P O / = -5 M; One fixed value. O * =,O =>,O = -5 / 9.47* -4 = -4 / / ( + ) - By trial and error, p = 8.. -p increased from 7.6 to 8.. his means O leaves the system to maintain a constant N.