ph calculations

Size: px

Start display at page:

Download "ph calculations"

Maurice Goodwin
5 years ago
Views:

1 Medical hemistry. Department of uman Physiology and Biochemistry p calculations Universal gas constant R= J/(mol K); Faraday s constant F=96485 /mol; T= K (human 37 ); T=98.15 K (standard conditions at room temperature 5 ). KEY EQUATINS: 1) p= lg [ ]; ) p= lg[ ]; 3) pp=14; 4) [ ]=10 p ; 5) [ ]=10 p ; 6) strong acids: [ ] = c M α z; 7) p = log [ ] = log(c M α z). S 4 S 4 divalent acid has two z= hydrogen cations 8) strong bases: 9) [ ] = c M α z; 10) p = log [ ] = log(c M α z). a() a divalent base has two z= hydroxyl ions; weak acids and bases K a dissociation constant of weak mono valent acid 3, salt (acetate 3 ); K b dissociation constant of weak mono valent base N 4, ammonium salt (acid N 4 ). [ ].[ 3 ] [ N 4 ].[ ] 11) K a = [ ] ; 1) K b = N nondis [ 4 ] ; 13) pk = lg[k]; 14) K = 10 pk ; nondis 3 15) [ ]= K a ; 16) [ pk ]= K b ; 17) p= a log pk ; 18) p= b log ; Amino acids AA have two type functional groups carboxylic and amino groups: [ AA ].[ ] [ AA N ].[ ] 19) K a = [ AA ] ; 0) K a = nondis [ AAN ] ; protonated 3 The carboxylic protolytic pair an acid a(aa) and conjugated basesalt b(aa ) AA AA, pk aaa < 5; acid a base b ; physiologic p=7.36 ranges between 5 < 7.36 < 8.5; AAN 3 AA N, pk aaan3 > 8.5 ; Protolytic pair acid a(aan 3 ) protonated N and conjugated base b AAN deprotonated N. smotic pressure π osm i isotonic coefficient;vant off s coefficient 1) π osm = i M R T; π osm = osm R T; M molar concentration mol/l; ) i = 1 α (m1); osm = i M osm osmomolar concentration mol/l; 3) [ ] = 10 p mol/l; R universal gas constant J/mol K; 4) [ ] = 10 p mol/l; T temperature K; K c α dissociation degree; 5) α= ; 6) α = dissoc m number of dissociated ions; c total S 4 => S 4 formed three ions in sum account m=1=3 ; a() a formed three ions in sum account m=1=3 ; free energsy change ΔG in driven process of concentration gradient 7) ΔG=RTln([ 3 citosol ]/[ 3 Mitohon ]), ΔG free energy change in driven process of bicarbonate concentration gradient [ 3 citosol ]/[ 3 Mitohon ] 8) ΔG=RTln([ 3 starpmembr]/[ 3 Mitohon]), ΔG positive free energy accumulation by oxidative phosphorylation driven protons concentration gradient [ 3 extramembr]/[ 3 Mitochon] 9) ΔG=RTln( outsm / insm ), ΔG free energy change by osmosis driven against concentration gradient outsm / insm make the pressure π=δ osm RT, where Δ osm = insm outsm E membr =Plog([ 3 extramit]/[ 3 Mitochon]); E 3 membr =Plog([ 3 cytosole ]/[ 3 Mitochon ]) 0) ΔG F = nf E membr_total, ΔG total membrane potential E membr E 3 membr free energy change [compound] molar concentration mol/l; osm osmo molar concentration mol/l; 1. To calculate p of the solution, that was obtained, when to 100 ml of l solution having p 1 =3 added 00 ml of water. M1 =10 3 M; M = M ; p =3.48. To V1 = 1 L liter solution of N 3 acid having p1 = 0.75 added V = 1 L liter solution of N 3 acid having p =. To calculate the p3 for obtained solution of strong acid solution! ( M1 = M ; M = 0.01 M ; M3 = M ; p 3 = 1.073) 1

2 Medical hemistry. Department of uman Physiology and Biochemistry 3. K dis = of a weak acid. To calculate the π osm osmotic pressure of these solution at 0, if the p = 1.5 of this solution! ( M =0.86 M; α = ; i = ; π osm = kpa) 4. K dis = of a weak base. To calculate the π osm osmotic pressure of these solution at 0, if the p = 10.5 of this solution! (p = 3.5 ; M = M; α = ; i = ; π osm = kpa) 5. alculate dissociation constant K a of Glutamate with concentration =0,100 mol/l and p=3,185! ompare it with average constant value pk a =(,199,674,5)/3=5,37... N R R, pk ar =4.5; αn 3 α N, pk an3 =9,67 ; α α, pk a =,19. At physiologic p=7, 36 ±0.01carboxylic groups negative charged and positive amino groups N 3 as pk reference to physiologic p value is smaller pk ar =4.5 < 7,36 and pk a =,19 <7,36 or greater 7,36 < 9,67= pk an3. 6. alculate dissociation constant K a of istidine with concentration =0,100 mol/l and p=3,3315! ompare it with average constant value pk a =(1,89,176,00)/3=5, N N R N N 3 R N, pk ar N =6.00; αn 3 α N, pk an3 =9,17 ; α α, pk a =1,8. 7. alculate dissociation constant K a of Aspartate with concentration =0,100 mol/l and p=3,015! ompare it with average constant value pk a =(1,889,603,65)/3=5, N R R, pk ar =3,65; αn 3 α N, pk an3 =9,60 ; α α, pk a =1, alculate dissociation constant K a of Lysine with concentration =0,100 mol/l and p=4,11! ompare it with average constant value pk a =(,188,9510,53)/3=7,... N N R N 3 R N, pk ar N3 =10.53; αn 3 α N, pk an3 =8,95 ; α α, pk a =, alculate p of asparagine with dissociation constant pk a =(,08,80)/=5,41; K a =10 5,41 pk and = 0,3 mol/l! p= a log =, p = and K dis = 10 4 of a weak acid. To calculate the volume V in ml of a weak acid that is necessary to take for neutralization of 0 ml 5 % Na sodium hydroxide. ρ = 1.04 g/cm 3. Macid =0.500 M; m Na =1.04 g; n Na =0.06 mol; MNa =1.30 M; Vacid=5 ml 11. p = 11. and K dis = of a weak base. ow many ml of 0.1 N l hydrochloric acid solution have to take for neutralization of 0 ml basic solution? ( Mbase =0.140 M; Vacid = ml) 1. ow much of Na sodium hydroxide in g grams have to solute into 500 ml water, that p of obtained solution would be p=1.54, α =90%, M Na =40g/mol. ( MNa =0.038 M; m Na =0.76g;) 13. p = 1.05 of l hydrochloric acid solution, dissociation degree α = 80 %. ow many ml are necessary to take for reaction by 10 ml 0. M K base solution? ( Macid α = N; Macid = M; n base = mmol/l;vacid = ml) 14. To calculate p of 1.85 % l hydrochloric acid solution, if dissociation degree α=80 %, and density of solution is 1 g/cm 3. M l =36.5 g/mol. M =0.507 M; p= smotic pressure of l hydrochloric acid solution at 0 temperature is 454 kpa, α=100 %. To calculate p of this solution! ( M = M; p=1.00 )

3 Medical hemistry. Department of uman Physiology and Biochemistry 16. p = 1 of sulfuric acid S 4 solution, dissociation degree α = 50 %. To calculate osmotic pressure at 0 temperature of this acidic solution! ( M =0.100M; π osm = kpa) 17. The five 5 liter of Na sodium hydroxide solution contain g Na. To find p of this solution, if density ρ =1.1g/cm3, α =1! ( M Na =40g/mol, n Na =0.05mol; MNa =0.01M; p=; p=1) 18. p = 1.5 of sulfuric acid S 4 solution, dissociation degree α = 80 %, density ρ = 1.05 g/cm 3. To calculate molarity and mass fraction! (M S4 = 98 g/mol ; M =0.0 M; 1.96%). 19. To calculate molarity of a() calcium hydroxide solution, if p = 13.. α = 90 %. ( M =0.088 M) 0. To calculate the p of 5 % formic acid solution. K dis., = 10 4, ρ = 1.0 g/cm 3. (m acid =51 g; M acid =46 g / mol; M = M; pk acid = 3.69; p = 1.87) 1. To calculate the p of 0.0 N 3 acetic acid solution. K dis,3 = (log = 1.7 ; p = 3.) 0. p = 0.7 of S 4 sulfuric acid solution, α=50 %, ρ=1.03 g/cm 3. To calculate mass fraction of this solution! M S4 = 98 g/mol! (log=0.7; w % =1.9%) 1. Extent 400mL of S 4 sulfuric acid solution contains 0.196g of S 4 of pure sulfuric acid. To calculate p and p for this solution! (M S4 =98g/mol, α =90%. (n S4 =0.00mol; M =0.005M;p=.045). p = 0 of S 4 sulfuric acid solution, α =70%. To calculate M and w %, if ρ =1.18g/mL! (1= M 0.7=1/1.4=0.714 M) 3. ow many times formic acid hydronium ions concentration [ 3 ] grater as the same concentration of acetic acid hydronium ions [ 3 ] concentration? K =1, ;K 3 =1, [ 3 ] concentration 3,18 times grater Student advanced self Studies (ome Work) exercises 4., 5., 6, 7. Physiologic solution mechanisms maintain balance 7.36=p, 3, in homeostasis. uman blood 7.36= p=log([ 3 ]), [ 3 ]= M = 10 p at stabilized arterial concentration [ aqua ]= M values prevent acidose, hypoxia (deficiency of oxygen) and oxidative stress by metabolism driving 3, 3, aqua concentration gradients across cell membranes through channels, 3,. 4. uman have p = 7.4 for blood and is p = 1. value for stomach juice. uman body between blood vessels and stomach cells exists as membranes separated equilibrium between two thermodynamics states for hydrogen ions in blood and in stomach: 3 blood <= Membrane => 3 stomach. K equilibrium =[ 3 ] stomach /[ 3 ] Blood ; ΔG r = R T ln(k equilibrium )=,3 R T log(k equilibrium ) What is the value of equilibrium constant? haracterize direction in which is equilibrium shifted? What is value of free energy change ΔG in this process of one mol 3 ion kj/mol? Which one direction of this flux process via membranes is spontaneous? [ 3 ] blood = M; [ 3 ] stomach = M;... K equilibrium = ;ΔG = kj/mol; spontaneous... 3

4 Medical hemistry. Department of uman Physiology and Biochemistry 5. alculate what is oxygen half reaction potential E if blood plasma hydrogen exponent is p = 7.36 and oxygen arterial concentration is [ aqua ] = M! At Faraday s constant F=96485 /mol ; R= J/(mol K); Why concentration in arterial blood avoid of non enzymatic xidation reactions? Why value of p=7.36 prevents acidosis, alkalosis and oxidative stress? uman, 3, shuttle oxy deoxy EMGLBIN controlled oxygen concentration [ aqua ]= M is formed from AIR 0.95% oxygen. Assuming, that human body temperature is t = 37 ; T = K and standard potential value Eo = V shows oxidizer oxygen very high power. Nernst s potential for half reaction equilibrium aqua 4 3 4e 6 in arterial blood plasma is: E =EoP/4log([ aqua ] [ 3 ] 4 ), where constant P in volts is P=,3 R T F = J ln(10) ( ) (K) mol K 96485( ) mol = V, E = /4*log( ( ) 4 )= *log( )=... = *log( )= * =... = = V;log( ( ) 4 )=log( )=... = *log( )= = V... From erythrocytes oxydeoxy, 3, shuttle hemoglobin equilibrium... reserves n = 0,057 M mol/l in one liter blood... restoring 459 times arterial based homeostasis concentration [ aqua ]= M.... Stabilization of aqua, 3, concentrations prevent destabilization created... acidose, alkalosis and oxidative stress as excess aqua and as deficiency hypoxia... as well as deviation of p 7.36 ±0.01 and oxygen concentration values... (arterial [ aqua ]= M and venous M) as decreased or increased... Potential arterial E = V and venous E = V below... standard value Eo = V;... Answer arterial : E = V and venous E = V 4

5 Medical hemistry. Department of uman Physiology and Biochemistry 6. What is total membrane potential E membr_total in volts? alculate the free energy change ΔG 3 for bicarbonate moving out of mitochondria in kj/mol! Physiological active Mitochondria have value of p = 7.36 inside and p = 5 in extra mitochondrial space. Bicarbonate concentration in cytosole and blood 0,0154 M: [ 3 ][ aqua ] = =0.03M. Using enderson aselbalh equation: p=7.36=pklog([ 3 cytosole ]/[ cytosole ]), ratio is alkaline reserve: =[ 3 cytosole ]/[ cytosole ]=.036/1= M/ M. As alkaline reserve in cytosole is known: [ 3 ]=.036*0.03/(1.036)= /3.036= M. In Mitochondria assuming M hemoglobin oxygen reserve used in Krebs cycle the total sum is [ 3 ][ aqua ]=0.03M0.057 M= 0,05054 M. As alkaline ratio keeps the same =[ 3 ]/[ aqua ]=.036/1: [ 3 ]=,036*0,05054/(1,036)=0,10899/3,036=0, M... Inside mitochondria bicarbonate concentration is two point two times greater.=[ 3 Mitochon ]/[ 3 citosols ]= M/ M. uman body temperature t=37 ; T = K; F=96485 /mol, n= 1 for 3... Actual membrane potential for bicarbonate anions with concentration gradient is: 3 Mitochon Membrane 3 cytosole Answer: E 3 Mitochon =Plog([ 3 cytosole ]/[ 3 Mitochon ]) =......=0,06154*log(0,0154/0, )=.0, V... Actual membrane potential for hydrogen ions 3 Mitochon Membrane 3 extramit... E membr =Plog([ 3 extramit]/[ 3 Mitochon])= log(10 5 / )= V Total membrane potential hydrogen and bicarbonate ions is : E membr = V = V; Membrane potential for bicarbonate, proton gradient as total membrane potential are: E 3 Mitochon =.0, V;... E membr =0, V;... E membr_total = ,1453= V;... Electric Free energy change for ion 3 negative charge n = 1 is: ΔG F = nf E membr_total = 1*96485* = kJ/mol.... oncentration gradient [ 3 citosol ]/[ 3 Mitochon ] made free energy decrease is: ΔG=RTln([ 3 citosol ]/[ 3 Mitochon ])= =8.3144*310,15*ln(0,0154/0, )= kJ/mol... Bicarbonate total free energy change ΔG 3 moving out of mitochondria is: ΔG 3 =ΔG F ΔG= (.0341) = kJ/mol 5

6 Medical hemistry. Department of uman Physiology and Biochemistry 7. Mitochondria Actual membrane potential for hydrogen cations 3 via the membrane proton channels and bicarbonate 3 channels reveal the equilibria: E membr membrane channels 3 membrane 3 Mitochon p= Mitochon proton cannel bicarbonate cannel 3 extramit p=5 3 cytosol [ 3 Mitochon ]=0,0339 M; [ 3 cytosol ]=0.015 M ydrogen and bicarbonate total membrane potential is sum of 0,1453V V = E membr =0,1663V. Electric free energy change for along the descendant potential 0,1663 V: ΔG E = E membr F n (ion charge 1) =0,1663*96485*(1) = 16,045 kj / mol G =RTln([ 3 ] extramit /[ 3 ] Mitochon )= =RTln(10 5 /10 7,36 )=8,3144*310,15*ln(10,36 )= 14,013 kj / mol free energy change for concentration gradient driven through proton channels crossing lipid bilayer membranes is the sum: ΔG membr = ΔG E G = 16,0454 kj / mol 14,013 kj / mol = 30,05846 kj / mol per one mole of proton drive ATPase to make work is 19 times per 3 effective as one mol mass one gram of proton in direction from extra membrane space ( 3 extramit ) to mitochondrial matrix space ( 3 Mitochon ). The proton concentration gradient ΔG= ΔG membr G channel = 30,058 kj/mol sum with electrochemical free energy change drive ATPase nano engine to synthesizing ATP molecules. Both free energy negative changes sum per one ATP mole is 4*30,058 kj/mol =10,3 kj/mol, consuming four protons 4, drive ATPase nano engine rotation to synthesizing one ATP molecule. ne mole 503 grams ATP production have been used 4 grams as four moles of protons. Free energy negative change is ΔG = 10 kj/mol. Macro ergic ATP phosphate anhydride bond in hydrolyze releases free energy ΔG = kj/mol for human erythrocyte ( page13). For ATP accumulated chemical free energy efficiency is 44,6% (53.47 kj/mol) of theoretically efficiency 100% (10 kj/mol). xidative phosphorylation 55,4% of used four proton chemo osmose energy consumes the friction of ATPase rotor to heat production and ATP movement in cytosol water medium forming the concentration gradients across lipid bilayer membranes as transportation free energy source to drive ATP molecules. Evidently any other charged cation molecule, for example, Na cation 3 times heavier or potassium cation K 39 times heavier and its relatively les efficiency per one gram of mass are transferred 3 times or 39 times less energy for ATP synthesis comparing with charged proton transfer through membrane channels. Life choose the best small by size, by mass and bearing whole one unit positive charge proton. 6

becomes the acid + b base

becomes the acid + b base IV. DISSCIATIN F WATER. p YDRGEN [ ] EXPNENT FUNCTIN. Āris Kaksis 016, Riga Stradin s University http://aris.gusc.lv/biothermodynamics/dissociation.doc Brensted protolytic concept of acid and base 193