C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (4): Acid-Base Eqm (4): B B uffer Solutions uffer Solutions Acid-Base Eqm (4): Acid-Base Eqm (4): B B uffer Solutions uffer Solutions Q.: Q.: What is the What is the new pH new pH after addition of after addition of 0.001 mol HCl at 298K? (assume no vol 0.001 mol HCl at 298K? (assume no vol ume change) ume change) 100 cm 100 cm 3 H H 2 O O 0.001 mol H 0.001 mol H Cl Cl [H [H 3 O O + ] ] = 0.00 0.00 1 100/100 100/100 0 = 0.010 M = 0.010 M pH = - pH = - log(0.010) = 2 log(0.010) = 2 pH would decrease pH would decrease from 7 to 2 from 7 to 2 after adding 0.001 mol HCl at 298K! after adding 0.001 mol HCl at 298K!
C. Y. Yeung (CHW, 2009)
Feb 03, 2016
Acid-Base Eqm (4): Buffer Solutions. Acid-Base Eqm (4): Buffer Solutions. p.01. 0.001 mol HCl. 0.001. = 0.010 M. [H 3 O + ] =. 100/1000. 100 cm 3 H 2 O. pH would decrease from 7 to 2 after adding 0.001 mol HCl at 298K!. - PowerPoint PPT Presentation
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C. Y. Yeung (CHW, 2009)
p.01Acid-Base Eqm (4):Acid-Base Eqm (4):
Buffer Solutions Buffer Solutions Acid-Base Eqm (4):Acid-Base Eqm (4):
Buffer Solutions Buffer Solutions
Q.:Q.: What is the What is the new pHnew pH after addition of 0.001 mol after addition of 0.001 mol HCl at 298K? (assume no volume change)HCl at 298K? (assume no volume change)
100 cm100 cm33 H H22OO
0.001 mol HCl0.001 mol HCl
[H[H33OO++] =] =0.0010.001
100/1000100/1000= 0.010 M= 0.010 M
pH = - log(0.010) = 2pH = - log(0.010) = 2
pH would decrease pH would decrease from 7 to 2from 7 to 2 after add after adding 0.001 mol HCl at 298K!ing 0.001 mol HCl at 298K!
p.02
Q.:Q.: What is the What is the new pHnew pH after addition of 0.001 mol after addition of 0.001 mol HCl at 298K? (assume no volume change)HCl at 298K? (assume no volume change)
100 cm100 cm33 0.1M CH 0.1M CH33COOH COOH KKaa = 1.76 = 1.761010-5-5MM
0.001 mol HCl0.001 mol HCl
pH would decrease pH would decrease from 2.88 to 1.99from 2.88 to 1.99 a after adding 0.001 mol HCl at 298K!fter adding 0.001 mol HCl at 298K!
BeforeBefore addition of HCl: addition of HCl: xx22
0.1 – x0.1 – x= 1.76= 1.761010-5-5
x = 1.33x = 1.331010-3-3
pH = 2.88pH = 2.88AfterAfter addition of HCl: addition of HCl:
extra [Hextra [H33OO++] = ] = 0.0010.001100/1000100/1000
= 0.01M= 0.01M
y(0.01+y)y(0.01+y)
0.1 – y0.1 – y= 1.76= 1.761010-5-5 y = 1.76y = 1.761010-4-4
pH = -log (0.01+y) = 1.99pH = -log (0.01+y) = 1.99
p.03
Q.:Q.: What is the What is the new pHnew pH after addition of 0.001 mol after addition of 0.001 mol HCl at 298K? (assume no volume change)HCl at 298K? (assume no volume change)
100 cm100 cm33 0.1M CH 0.1M CH33COOH COOH and 0.1M CHand 0.1M CH33COOCOO-- Na Na++
0.001 mol HCl0.001 mol HCl
pH would decrease pH would decrease from from 4.754.75 to to 4.674.67 a after adding 0.001 mol HCl at 298K!fter adding 0.001 mol HCl at 298K!
BeforeBefore addition of HCl: addition of HCl: x (0.1+x)x (0.1+x)0.1 – x0.1 – x
= 1.76= 1.761010-5-5
x = 1.76x = 1.761010-5-5
pH = 4.75pH = 4.75
AfterAfter addition of HCl: extra [H addition of HCl: extra [H33OO++] = 0.01M] = 0.01M
[CH[CH33COOCOO--] remained = 0.1 – 0.01 = 0.09M] remained = 0.1 – 0.01 = 0.09M
(0.09+y) y(0.09+y) y
0.11 – y0.11 – y= 1.76= 1.761010-5-5 y = 2.15y = 2.151010-5-5
pH = -log (y) = 4.67pH = -log (y) = 4.67
p.04
Note the difference …Note the difference …
100cm3 H2O
100cm3 0.1M CH3COO
H
100cm3 0.1M CH3COOH and 0.1M CH3
COO-Na+
Initial pH 7 2.88 4.75pH after adding 0.001 mol HCl 2 1.99 4.67
tends to resist the tends to resist the changes in pH changes in pH when small amounts of acid is addedwhen small amounts of acid is added
i.e. it is a “BUFFER” solution!i.e. it is a “BUFFER” solution!
p.05
100 cm100 cm33 0.1M CH 0.1M CH33COOH COOH and 0.1M CHand 0.1M CH33COOCOO-- Na Na++
It is an ACIDIC It is an ACIDIC BUFFERBUFFER solution! solution! (i.e. weak acid + conjugate base)(i.e. weak acid + conjugate base)
It tends to resist changes in pH when It tends to resist changes in pH when small amounts of acid or base are added, small amounts of acid or base are added, and their pH is not affected by dilution.and their pH is not affected by dilution.
When When acid is addedacid is added, the additional H, the additional H33OO++ would combine with the conjugate base would combine with the conjugate base so that the eqm shifts BW.so that the eqm shifts BW.i.e. i.e. [H[H33OO++] does not change much, ] does not change much,
pH remains almost unchanged.pH remains almost unchanged.
When When base is addedbase is added, some H, some H33OO++ ions are r ions are reacted. Some acid molecules dissociate teacted. Some acid molecules dissociate to produce Ho produce H33OO++ so that the eqm shifts FW. so that the eqm shifts FW.i.e. i.e. [H[H33OO++] does not change much, ] does not change much,
pH remains almost unchanged.pH remains almost unchanged.
CHCH33COOHCOOH + + HH22OO CHCH33COOCOO-- + + HH33OO++
p.06
100 cm100 cm33 0.1M NH 0.1M NH33 and and 0.1M NH0.1M NH44ClCl
It is an BASIC It is an BASIC BUFFERBUFFER solution! solution! (i.e. weak base + conjugate acid)(i.e. weak base + conjugate acid)
When When acid is addedacid is added, some OH, some OH-- are neutr are neutralized, some NHalized, some NH33 molecules would disso molecules would dissociate so that the eqm shifts FW.ciate so that the eqm shifts FW.i.e. i.e. [OH[OH--] does not change much, pH ] does not change much, pH
remains almost unchanged.remains almost unchanged.
When When base is addedbase is added, additional OH, additional OH-- ions w ions would combine with NHould combine with NH44
++ so that the eqm sh so that the eqm shifts BW.ifts BW.i.e. i.e. [OH[OH--] does not change much, pH ] does not change much, pH
remains almost unchanged.remains almost unchanged.
NHNH33 + + HH22OO NHNH44++ + + OHOH--
p.07
pH of buffer is NOT affected by dilution.pH of buffer is NOT affected by dilution. WHY?WHY?
pKpKaa = =[A[A--(a(a
q)]q)][HA(aq)][HA(aq)]- log- logpHpH
conjugate baseconjugate base
acidacidconstant at constant at constant temp.constant temp.
During dilution, both [ADuring dilution, both [A--] and [HA] are ] and [HA] are diluted to the diluted to the same extentsame extent. Thus the ratio [A. Thus the ratio [A--]/[HA] is unchanged.]/[HA] is unchanged.
Therefore, pH of buffer remains unchanged in dilution!Therefore, pH of buffer remains unchanged in dilution!
pH of the buffer could be adjusted pH of the buffer could be adjusted by varying the ratio of HA and Aby varying the ratio of HA and A--..
p.08Q.1:Q.1: Lactic acid (Lactic acid ( 乳酸乳酸 ), CH), CH33CH(OH)COOH, is produced in musCH(OH)COOH, is produced in muscle tissues during physical exertion. cle tissues during physical exertion.
At body temperature, human blood pH is 7.40. At body temperature, human blood pH is 7.40.
Whether the lactic acid produced exists mainly in form of Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociated lactate ions” in “undissociated molecule” or “dissociated lactate ions” in blood at body temperature?blood at body temperature?(Given: K(Given: Kaa of lactic acid at body temperature = 8.50 of lactic acid at body temperature = 8.501010-4-4 mol dm mol dm--
33))
pKpKaa = =[CH[CH33CH(OH)COOCH(OH)COO--]][CH[CH33CH(OH)COOH]CH(OH)COOH]
pH –pH – loglog
2.14 2.14 10104 4 ==[CH[CH33CH(OH)COOH]CH(OH)COOH][CH[CH33CH(OH)COOCH(OH)COO--]]
Lactic acid formed is mainly in form of Lactic acid formed is mainly in form of dissociated lactic ionsdissociated lactic ions. .
With a given Ka, the ratio of [salt] and [acid] could be found from the pH value.
p.09Q.2:Q.2: Calculate the pH value of the resultant solution for Calculate the pH value of the resultant solution for
10cm10cm33 of 0.20M CH of 0.20M CH33COOH are mixed with 10cmCOOH are mixed with 10cm33 of of 0.40M CH0.40M CH33COONa. (KCOONa. (Kaa = 1.75 = 1.751010-5-5 mol dm mol dm-3-3))
x(0.20+x)x(0.20+x)0.10 – x0.10 – x
1.751.751010-5-5 = =
x = 8.75 x = 8.75 1010-6-6
pH = 5.06pH = 5.06
After mixing,After mixing, new [CHnew [CH33COOH] = 0.10MCOOH] = 0.10Mnew [CHnew [CH33COONa] = 0.20MCOONa] = 0.20M
An acid-salt mixture (acid and salt have An acid-salt mixture (acid and salt have comparablecomparable conc.) conc.)
ACIDIC BUFFER! ACIDIC BUFFER!
pH of pH of acidicacidic buffer buffer could be adjusted by could be adjusted by [acid] and [salt].[acid] and [salt].
p.10Q.3:Q.3: Calculate the pH value of the resultant solution for Calculate the pH value of the resultant solution for
10cm10cm33 of 1.00M NH of 1.00M NH33 are mixed with 10cm are mixed with 10cm33 of 1.00M of 1.00M NHNH44Cl. (KCl. (Kbb = 1.78 = 1.781010-5-5 mol dm mol dm-3-3))
x(0.50+x)x(0.50+x)0.50 – x0.50 – x
1.781.781010-5-5 = =
x = 1.78 x = 1.78 1010-5-5
pOH = 4.75pOH = 4.75
After mixing,After mixing, new [NHnew [NH33] = 0.50M] = 0.50Mnew [NHnew [NH44
++] = 0.50M] = 0.50M
A base-salt mixture (base and salt have A base-salt mixture (base and salt have comparablecomparable conc.) conc.)
BASIC BUFFER! BASIC BUFFER!
pH = 9.25pH = 9.25
pH of pH of basicbasic buffer buffer
could be adjusted by
could be adjusted by [base] and [salt].
[base] and [salt].
p.11
Article Reading … about Medical Science!Article Reading … about Medical Science!
Acid-Base Equilibria & pH Regulation iAcid-Base Equilibria & pH Regulation in Blood Plasma n Blood Plasma
AssignmentAssignmentp.12
Next ….Next ….Acid-Base Indicators (Book 2 p. 164 Acid-Base Indicators (Book 2 p. 164 – 168)– 168)
Study the examples in p. 158 - 163,Study the examples in p. 158 - 163,
p.163 p.163 Check point 18.2 , Check point 18.2 ,
p.179 p.179 Q.1 - 6Q.1 - 6
[due date: 23/4(Thur)] [due date: 23/4(Thur)]
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