BUFFERSOLUTION

1、department of medical chemistryyongxiang hu1. what is a buffer solutionexample: in 1 liter pure water add 0.01 mol of hydrochloric acid. calculate this solutions ph change value. and in 1 liter contain 0.1 mol acetic acid and 0.1 acetate ion mixture solution add 0.01mol of hydrochloric acid. calcula

2、te this solutions ph change value. notice: square bracket will instead of equilibrium concentration in following time. such as h+ instead hydrogen ion equilibrium concentration.step 1. calculate the pure waters ph change value the pure waters hydrogen ion exponent (ph) value: ph(pure water) = 7.00wh

3、en add the hydrochloric acid: h+ = 0.01mol/1 liter =0.01mol.l-1 ph = -log 0.01 = 2.00 ph = 7.00 2.00 = 5.00 (ph unit) step 2. calculate contain acetic acid and its conjugate base solutions ph value in the solution acetic acids concentration and acetate ions concentration hac = 0.10 mol.l-1 ac- = 0.1

4、0 mol.l-1 acetic acid equilibrium-constant ka = 1.7010-5 in this solution has a acid-base equilibrium hac h+ + ac- starting 0.10 0.10substitute the concentrations into the equilibrium-constant equation ka = h+ac-/hac h+ = ka hac/ac- = 1.70 10-5 0.10/ 0.10 = 1.70 10-5 (mol.l-1) ph = 4.77step 3. calcu

5、late solutions ph value when add hydrochloric acid hac h+ + ac- starting 0.10 0.10 change +0.01 -0.01equilibrium 0.10+0.01 0.10-0.01step 4. substitute the equilibrium concentration into the equilibrium-constant equation h+ = ka hac/ac- = 1.70 10-5 0.11/ 0.09 = 2.07 10-5 ph = 4.68 ph = 4.774.68= 0.09

6、 (ph unit) definition of buffer solution : a buffer solution is a solution characterized by the ability to resist changes in ph when limited amounts of strong acid or strong base are added to it.2. composition of buffer solution buffer solution consist of weak acid and its conjugate base or weak bas

7、e and its conjugate acid. the conjugate acid-base pair are called buffer system or buffer pair. important buffer pair such as: hac - ac- nh4+ - nh3 h2co3 - hco3- h3po4 - h2po4- h2po4- - hpo42- hpo42- - po43-3. calculate the ph value of buffer solutionhenderson-hasselbalch equation:a buffer solution

8、made up of a weak acid ha and its conjugate base a-. the acid-ionization equilib-rium is ha(aq) h+(aq) + a-(aq)and the acid-ionization constant is ka = h+a-/haaccording to the constant equation, we can get anequation for the h+ concentration. h+ = ka ha/a-take the negative logarithm of both sides of

9、 the equation. that is, -logh+ = -logka ha/a- = -log ka log ha/a-the left side equals the ph. the pka of a weak acid is defined in a manner similar to ph and poh. pka = -log ka the previous equation can be written ph = pka - log ha/a- = pka + log a-/hamore generally, we can write base ph = pka + log

10、 acidthis is known as the henderson-hasselbalch equation.buffer ratio: base/acid hendersonhasselbalch equation change form: phpka+lg pka+lg hbbc(conjugate base)c(conjugate acid)phpka+lg pka+lg vnvn/ )hb(/ )b()hb()b(nnphpka+lg pka+lg )hb()hb()b()b(vcvc)hb()b(vvbuffer capacity and buffer effective ran

11、ge1. buffer capacity() is definited as the amount of substance of strong acid or strong base when the ph value change one unit of ph unit in 1 liter of buffer solution. with differential coefficient: = dph d)b(avn 2.303hbb-/ctotal dph d)b(avn(1)(2) 0.1mol.l-1 hac+ naoh(3) 0.2mol.l-1 hac+naoh(4)phmax

12、2.303(ct /2)(ct /2) /ct 0.576ct cthb+b- 2.303hbb-/ct 2.303(hb /ct )(b-/ct )ct 2.303(hb /ct )(1-hb-/ct )ct2. buffer effective range when the buffer ratio is 110 or 101, we cognize this buffer solution without the buffer ability. so, we can definite the buffer effective range as:ph = pka1prepare the b

13、uffer solution principle and step 1. choose correspondence of buffer system2. the total concentration of buffer solution3. calculate the amount of buffer solution4. calibrate the buffer solutiontris: trihydrixymethylaminomethane(hoch2)3cnh2 standard of buffer solutionpotassium hydrotartrate (khc4h4o

14、6) potassium acid phthalate (khc8h4o4) borax (na2b4o710h2o) phosphate (kh2po4-na2hpo4) buffer system in bloodin plasma:h2co3-hco3-、h2po4-hpo42-、hnp-hn-1p-(hnp instead of protein) in red cell: h2b-hb- (h2b instead of haemoglobin)、h2bo2-hbo2- (h2bo2 instead of oxyhemoglobin)、h2co3-hco3-、h2po4-hpo42- c

15、o2(solute)+h2o h2co3 h+ hco3-phpka，+lghco3-co2solutehco3-co2solute 6.10 + lg in plasma:hco3- = 0.024moll-1co2solute= 0.001 2moll-1 ph6.10 + lg 6.10 + lg 7.4011lmol0012. 0lmol024. 0120summa1.buffer solution is composed of a pair of conjugate acid-base pair with enough concentration.the component of a

16、nti-acid is conjugate base. the component of anti-base is conjugate acid. when a few strong acid or base is added in to the buffer solution.the component of anti-acid and anti -base will move the equilibrium of dissociation to stabilize ph value.2. we can use the henderson-hasselbalchs equation to c

17、alculate the ph value of buffer solution: phpka+lg pka+lg pka+lg hbb)hb()b(cc)hb()b(nn3. buffer capacity is decided by pan-concentration and buffer ratio. when the pan-concentration of the buffer pair is settled, the buffer ratio is the more close with to 1 , the buffer capacity will be the more lar

18、ge. when the buffer ratio is equal to 1, the buffer capacity will attain the maximum max. when the buffer ratio is equalization, the pan-concentration of the buffer pair is the more large, the buffer capacity will be the more large.4. phpka土1 is stipulated the buffer effective range. over this range, the buffer solution will be to lose the buffer ability too low of the buffer capacity. dispense the buffer solution. it may be according to the table of this