11

Intravenous Infusion Equations Intravenous Infusion Equations Some drugs are administered using a continuous intravenous Some drugs are administered using a continuous intravenous

infusion, and if the infusion is discontinued the serum infusion, and if the infusion is discontinued the serum concentration/time profile decreases in a straight line when concentration/time profile decreases in a straight line when graphed on a semilogarithmic axes. In this case, a one graphed on a semilogarithmic axes. In this case, a one compartment model intravenous infusion equation can be used compartment model intravenous infusion equation can be used to compute concentrations (C) while the infusion is running: C to compute concentrations (C) while the infusion is running: C =(k=(k00/Cl)(1 − e /Cl)(1 − e −ket−ket) =[k) =[k00/(k/(keeVVdd)](1 −e )](1 −e −ket−ket), where k), where k00 is the drug is the drug infusion rate (in amount per unit time, such as mg/h or infusion rate (in amount per unit time, such as mg/h or μg/min), Cl is the drug clearance (since Cl =kμg/min), Cl is the drug clearance (since Cl =kelelVVdd , this , this substitution was made in the second version of the equation), substitution was made in the second version of the equation), kkelel is the elimination rate constant, and t is the time that the is the elimination rate constant, and t is the time that the infusion has been running. infusion has been running.

22

If a drug is given as a continuous If a drug is given as a continuous intravenous infusion, serum intravenous infusion, serum concentrations increase until a concentrations increase until a steady-state concentration steady-state concentration (Css) (Css) is is achieved in 5–7 half-lives. The achieved in 5–7 half-lives. The steady-state concentration is steady-state concentration is determined by the quotient of the determined by the quotient of the infusion rate infusion rate (k(k00) ) and drug and drug

clearance clearance (Cl): (Cl): Css = kCss = k00/Cl. When /Cl. When

the infusion is discontinued, serum the infusion is discontinued, serum concentrations decline in a straight concentrations decline in a straight line if the graph is plotted on line if the graph is plotted on semilogarithmic axes. When using semilogarithmic axes. When using log10 graph paper, the elimination log10 graph paper, the elimination rate constant (krate constant (kelel) can be computed ) can be computed

using the following formula: slope using the following formula: slope =−k=−kelel/2.303./2.303.

33

If the infusion is stopped, postinfusion serum If the infusion is stopped, postinfusion serum concentrations (Cconcentrations (Cpostinfusionpostinfusion) can be computed by ) can be computed by calculating the concentration when the infusion calculating the concentration when the infusion ended (Cended (Cendend) using the appropriate equation in ) using the appropriate equation in the preceding paragraph, and the following the preceding paragraph, and the following equation: Cequation: Cpostinfusionpostinfusion = C = Cendendee−kel*t(postinfusion)−kel*t(postinfusion), where k, where kelel is the elimination rate constant and tis the elimination rate constant and tpostinfusionpostinfusion is is the postinfusion time (tthe postinfusion time (tpostinfusionpostinfusion = 0 at end of = 0 at end of infusion and increases from that point). infusion and increases from that point).

44

A. Constant rate infusionA. Constant rate infusion

Cp = [k0/(keVd)](1 − e−ket) Where k0 is the infusion rate.

The plasma concantration at any time during the infusion can be calculated from the relation:

Cp = [Dose/(keVd*Tinfusion)](1 − e−ket)Where k0 = Dose/Tinfusion

55

b. Single dose infusionb. Single dose infusion

For a single infusion dose, the peak plasma For a single infusion dose, the peak plasma concentration can be calculated from the concentration can be calculated from the relation:relation:

CCp, 1(max)p, 1(max) = {Dose/(K = {Dose/(Kelel*V*Vdd*T)} [1 - e*T)} [1 - e-Kel*T-Kel*T]]

Where T is the infusion timeWhere T is the infusion time

66

c. Multiple dose infusionc. Multiple dose infusion When multiple doses are administered by infusion, When multiple doses are administered by infusion,

the maximum plasma concentration can be the maximum plasma concentration can be determined using the eruation:determined using the eruation:

CCp(max)p(max)={Dose/(K={Dose/(Kelel*V*Vdd*T)}[(1 - e*T)}[(1 - e-Kel*T-Kel*T)/(1 - e)/(1 - e-Kel*-Kel*]]

In the same manner, the minimum plasma In the same manner, the minimum plasma concentration after multiple infusions can be concentration after multiple infusions can be determined using the erqation:determined using the erqation:

CCp(min)p(min) = C = Cp(max)p(max) e e-Kel*(-Kel*( – T) – T)

Where Where is the infusion interval is the infusion interval

77

d. Infusion at steady stated. Infusion at steady state

After a long time of infusionAfter a long time of infusion (1 − e−ke*t) = 1, and thus we have: CCp,ssp,ss = k = k00/(K/(Kelel*V*Vdd)) After termination of infusion, we have:After termination of infusion, we have:

CCp(end)p(end)= C= Cppoo and the first order kinetics equation leads to: and the first order kinetics equation leads to:

CCpp = C = Cp(end)p(end) e e-Kel*t-Kel*t

CCp(end)p(end) = {k = {koo/(K/(Kelel*V*Vdd)} [1 - e)} [1 - e-Kel*t(infusion)-Kel*t(infusion)]]

Where CWhere Cp(end)p(end) is the plasma concentration when the is the plasma concentration when the infusion is stopped infusion is stopped

88

Other ParametersOther Parameters

99

For example, a patient is administered 60 mg/h of For example, a patient is administered 60 mg/h of theophylline. It is known from previous hospital theophylline. It is known from previous hospital admissions that the patient has the following admissions that the patient has the following pharmacokinetic parameters for theophylline: V = 40 pharmacokinetic parameters for theophylline: V = 40 L and kL and kelel = 0.139 h = 0.139 h−1−1. The serum concentration of . The serum concentration of

theophylline in this patient after receiving the drug theophylline in this patient after receiving the drug for 8 hours and at steady state can be calculated: C = for 8 hours and at steady state can be calculated: C = [k[k00/(k/(kelelV)](1 − e V)](1 − e −ket−ket) = [(60 mg/h)/(0.139 h) = [(60 mg/h)/(0.139 h−1−1 40 L)]⋅ 40 L)]⋅(1 − e (1 − e −(0.139 h−1)(8 h)−(0.139 h−1)(8 h) ) = 7.2 mg/L; Css = k ) = 7.2 mg/L; Css = k00/(k/(kelelVVdd) = (60 ) = (60

mg/h)/(0.139 hmg/h)/(0.139 h−1−1 40 L) = 10.8 mg/L. ⋅ 40 L) = 10.8 mg/L. ⋅

1010

It is possible to compute the theophylline serum It is possible to compute the theophylline serum concentration 6 hours after the infusion stopped in concentration 6 hours after the infusion stopped in either circumstance. If the infusion only ran for 8 either circumstance. If the infusion only ran for 8 hours, the serum concentration 6 hours after the hours, the serum concentration 6 hours after the infusion stopped would be: Cinfusion stopped would be: Cpostinfusionpostinfusion = = CCendendee−ke*t(postinfusion)−ke*t(postinfusion) = (7.2 mg/L) e = (7.2 mg/L) e −(0.139 h−1)(6 h)−(0.139 h−1)(6 h) = 3.1 = 3.1 mg/L. If the infusion ran until steady state was mg/L. If the infusion ran until steady state was achieved, the serum concentration 6 hours after the achieved, the serum concentration 6 hours after the infusion ended would be: Cinfusion ended would be: Cpostinfusionpostinfusion = C = Cendendee−ke*t(postinfusion)−ke*t(postinfusion) = (10.8 mg/L) e= (10.8 mg/L) e−(0.139 h−1)(6 h)−(0.139 h−1)(6 h) = 4.7 mg/L. = 4.7 mg/L.

1111

Even if serum concentrations exhibit a Even if serum concentrations exhibit a distribution phase after the drug infusion has distribution phase after the drug infusion has ended, it is still possible to use one ended, it is still possible to use one compartment model intravenous infusion compartment model intravenous infusion equations for the drug without a large amount equations for the drug without a large amount of error. The strategy used in this instance is to of error. The strategy used in this instance is to infuse the medication and wait for the infuse the medication and wait for the distribution phase to be over before measuring distribution phase to be over before measuring serum drug concentrations in the patient. serum drug concentrations in the patient.

1212

For example, gentamicin, tobramycin, and amikacin For example, gentamicin, tobramycin, and amikacin are usually infused over one-half hour. When are usually infused over one-half hour. When administered this way, these aminoglycoside administered this way, these aminoglycoside antibiotics have distribution phases that last about antibiotics have distribution phases that last about one-half hour. Using this strategy, aminoglycoside one-half hour. Using this strategy, aminoglycoside serum concentrations are obtained no sooner than serum concentrations are obtained no sooner than one-half hour after a 30-minute infusion in order to one-half hour after a 30-minute infusion in order to avoid the distribution phase. If aminoglycosides are avoid the distribution phase. If aminoglycosides are infused over 1 hour, the distribution phase is very infused over 1 hour, the distribution phase is very short and serum concentrations can be obtained short and serum concentrations can be obtained immediately. immediately.

1313

For example, a patient is given an intravenous For example, a patient is given an intravenous infusion of gentamicin 100 mg over 60 minutes. infusion of gentamicin 100 mg over 60 minutes. Because the patient received gentamicin before, it is Because the patient received gentamicin before, it is known that the volume of distribution is 20 L, the known that the volume of distribution is 20 L, the elimination rate constant equals 0.231 helimination rate constant equals 0.231 h−1−1, and the , and the half-life equals 3 h (thalf-life equals 3 h (t1/21/2 = 0.693/k= 0.693/0.231 h = 0.693/k= 0.693/0.231 h−1−1 = = 3 h). To compute the gentamicin concentration at the 3 h). To compute the gentamicin concentration at the end of infusion, a one compartment model end of infusion, a one compartment model intravenous infusion equation can be employed: intravenous infusion equation can be employed:

C = [kC = [k00/(k/(kelelVVdd)](1 − e)](1 − e−kelt−kelt) = [(100 mg/1 h)/ (0.231 h) = [(100 mg/1 h)/ (0.231 h−1−1 ⋅ ⋅20 L)](1 − e20 L)](1 − e−(0.231 h−1)(1 h)−(0.231 h−1)(1 h)) = 4.5 mg/L. ) = 4.5 mg/L.

1414

Pharmacokinetic constants can also be calculated for Pharmacokinetic constants can also be calculated for use in the equations. If a steady-state concentration is use in the equations. If a steady-state concentration is obtained after a continuous intravenous infusion has obtained after a continuous intravenous infusion has been running uninterrupted for 3–5 half-lives, the been running uninterrupted for 3–5 half-lives, the drug clearance (Cl) can be calculated by rearranging drug clearance (Cl) can be calculated by rearranging the steady-state infusion formula: Cl = kthe steady-state infusion formula: Cl = k00/Css. For /Css. For example, a patient receiving procainamide via example, a patient receiving procainamide via intravenous infusion (kintravenous infusion (k00 = 5 mg/min) has a steady- = 5 mg/min) has a steady-state procainamide concentration measured as 8 state procainamide concentration measured as 8 mg/L. Procainamide clearance can be computed using mg/L. Procainamide clearance can be computed using the following expression: Cl = kthe following expression: Cl = k00/Css = (5 mg/min) / /Css = (5 mg/min) / (8 mg/L) = 0.625 L/min. (8 mg/L) = 0.625 L/min.

1515

If the infusion did not run until steady state was If the infusion did not run until steady state was achieved, it is still possible to compute achieved, it is still possible to compute pharmacokinetic parameters from postinfusion pharmacokinetic parameters from postinfusion concentrations. In the following example, a patient concentrations. In the following example, a patient was given a single 120-mg dose of tobramycin as a was given a single 120-mg dose of tobramycin as a 60-minute infusion, and concentrations at the end of 60-minute infusion, and concentrations at the end of infusion (6.2 mg/L) and 4 hours after the infusion infusion (6.2 mg/L) and 4 hours after the infusion ended (1.6 mg/L) were obtained. By plotting the ended (1.6 mg/L) were obtained. By plotting the serum concentration/time information on serum concentration/time information on semilogarithmic axes, the half-life can be determined semilogarithmic axes, the half-life can be determined by measuring the time it takes for serum by measuring the time it takes for serum concentrations to decline by one-half, and equals 2 concentrations to decline by one-half, and equals 2 hours in this case. hours in this case.

C = [k0/(keVd)](1 − e−ket)

1616

1717

The elimination rate constant (kThe elimination rate constant (kee) can be calculated using the ) can be calculated using the following formula: following formula:

kkee = 0.693/t = 0.693/t1/21/2 =0.693/2 h =0.347 h =0.693/2 h =0.347 h−1−1. .

Alternatively, the elimination rate constant can be calculated Alternatively, the elimination rate constant can be calculated without plotting the concentrations using the following without plotting the concentrations using the following equation: equation:

kkee = −(ln C = −(ln C11 − ln C − ln C22)/(t)/(t11 −t −t22), ),

where twhere t11 and C and C11 are the first time/concentration pair and t are the first time/concentration pair and t22 and C and C22 are the second time/concentration pair; are the second time/concentration pair;

kkee = −[ln (6.2 mg/L) − ln (1.6 mg/L)] / (1 h −5 h) = 0.339 h = −[ln (6.2 mg/L) − ln (1.6 mg/L)] / (1 h −5 h) = 0.339 h−1−1 (note the slight difference in k(note the slight difference in kee is due to rounding errors). The is due to rounding errors). The elimination rate constant can be converted into the half-life elimination rate constant can be converted into the half-life using the following equation: using the following equation:

tt1/21/2 =0.693/k =0.693/kee = 0.693/0.339 h = 0.693/0.339 h−1−1 =2 h. =2 h.

1818

1919

Drug AccumulationDrug Accumulation

With repeating drug doses, the drug will accumulate With repeating drug doses, the drug will accumulate in the body until dosing ceasesin the body until dosing ceases. .

PracticallyPractically: : accumulation will be observed if the accumulation will be observed if the dosing interval is less than 4 halfdosing interval is less than 4 half--liveslives. .

AccumulationAccumulation: : inversely proportional to the fraction inversely proportional to the fraction of the dose lost in each dosing intervalof the dose lost in each dosing interval Accumulation factor Accumulation factor = = 11//Fraction lost in one dosing Fraction lost in one dosing

interval interval = = 11/(/(1 1 - - fraction remainingfraction remaining( ( For example, the accumulation factor for a drug given For example, the accumulation factor for a drug given

once every halfonce every half--lifelife: : 1/0.5 equals 21/0.5 equals 2. .

2020

Accumulation FactorAccumulation Factor

The ratio, Dose/V represents the initial The ratio, Dose/V represents the initial concentration after the first dose (Cpconcentration after the first dose (Cp11

oo) . Thus ) . Thus

the ratio between the highest, initial the ratio between the highest, initial concentration at steady state, Cconcentration at steady state, Cpmaxpmax and the and the

highest concentration after the first dose, Cphighest concentration after the first dose, Cp0011

can be expressed as the ratio: can be expressed as the ratio: R equal to e R equal to e -kel•-kel•

2121

Loading doseLoading dose

The time required to obtain steady-state plasma The time required to obtain steady-state plasma levels by IV infusion will be long. It is levels by IV infusion will be long. It is possible to administer an intravenous loading possible to administer an intravenous loading dose to attain the desired drug concentration dose to attain the desired drug concentration immediately and then attempt to maintain this immediately and then attempt to maintain this concentration by a continuous infusion.concentration by a continuous infusion.

2222

2323

2424

If the loading dose is to be administered orally, then If the loading dose is to be administered orally, then the bioavailability term (the bioavailability term (FF) needs to be introduced. ) needs to be introduced. ThusThus::

2525

A 100 mg dose was administered every 8 h. At A 100 mg dose was administered every 8 h. At steady state, two plasma concentrations are steady state, two plasma concentrations are measured:measured:

Sample 1 is taken at 1 h post dose: Conc Sample 1 is taken at 1 h post dose: Conc 9.6 9.6 mg/Lmg/L

Sample 2 is taken pre dose: Conc Sample 2 is taken pre dose: Conc 2.9 mg/L2.9 mg/L

2626

2727

2828

2929

The The volume of distribution (Vvolume of distribution (Vdd)) can be can be

calculated from either the 1 h post- or pre-dose calculated from either the 1 h post- or pre-dose samples. From the 1 h post-dose sample samples. From the 1 h post-dose sample

The following equation describes the plasma The following equation describes the plasma concentration 1 h post dose at steady state:concentration 1 h post dose at steady state:

3030

3131

3232

3333

Extravascular EquationsExtravascular Equations When a drug is administered extravascularly (e.g., orally, When a drug is administered extravascularly (e.g., orally,

intramuscularly, subcutaneously, transdermally, etc.), intramuscularly, subcutaneously, transdermally, etc.), absorption into the systemic vascular system must take place. absorption into the systemic vascular system must take place. If serum concentrations decrease in a straight line when If serum concentrations decrease in a straight line when plotted on semilogarithmic axes after drug absorption is plotted on semilogarithmic axes after drug absorption is complete, a one compartment model extravascular equation complete, a one compartment model extravascular equation can be used to describe the serum concentration/time curve: can be used to describe the serum concentration/time curve:

C = {(FkC = {(FkaaD) /[V(kD) /[V(kaa −k −kee)]}(e)]}(e−ket−ket −e −e−kat−kat), ), where t is the time after the extravascular dose was given (t =0 where t is the time after the extravascular dose was given (t =0

at the time the dose was administered), C is the concentration at the time the dose was administered), C is the concentration at time =t, F is the bioavailability fraction, kat time =t, F is the bioavailability fraction, kaa is the absorption is the absorption rate constant rate constant

3434

The absorption rate constant The absorption rate constant (k(kaa) ) controls how quickly the drug controls how quickly the drug enters the body. A large enters the body. A large absorption rate constant allows absorption rate constant allows drug to enter the body quickly drug to enter the body quickly while a small absorption rate while a small absorption rate constant permits drug to enter the constant permits drug to enter the body more slowly. The body more slowly. The solid line solid line shows the concentration/time shows the concentration/time curve on semilogarithmic axes for curve on semilogarithmic axes for an elimination rate constant equal an elimination rate constant equal to 2 hto 2 h−1−1. The . The dashed dashed and and dotted dotted lines lines depict serum depict serum concentration/time plots for concentration/time plots for elimination rate constants of 0.5 elimination rate constants of 0.5 hh−1−1 and 0.2 h and 0.2 h−1−1, respectively. , respectively.

3535

An example of the use of this equation would An example of the use of this equation would be a patient that is administered 500 mg of oral be a patient that is administered 500 mg of oral procainamide as a capsule. It is known from procainamide as a capsule. It is known from prior clinic visits that the patient has a half-life prior clinic visits that the patient has a half-life equal to 4 hours, an elimination rate constant equal to 4 hours, an elimination rate constant of 0.173 hof 0.173 h−1−1 (k= 0.693/t (k= 0.693/t1/21/2 = 0.693/4 h = 0.173 = 0.693/4 h = 0.173 hh−1−1), and a volume of distribution of 175 L. ), and a volume of distribution of 175 L. The capsule that is administered to the patient The capsule that is administered to the patient has an absorption rate constant equal to 2 hhas an absorption rate constant equal to 2 h−1−1, , and an oral bioavailability fraction of 0.85. and an oral bioavailability fraction of 0.85.

3636

The procainamide serum concentration 4 hours The procainamide serum concentration 4 hours after a single dose would be equal to: after a single dose would be equal to:

3737

If the serum concentration/time curve displays If the serum concentration/time curve displays a distribution phase, it is still possible to use a distribution phase, it is still possible to use one compartment model equations after an one compartment model equations after an extravascular dose is administered. In order to extravascular dose is administered. In order to do this, serum concentrations are obtained only do this, serum concentrations are obtained only in the postdistribution phase. Since the in the postdistribution phase. Since the absorption rate constant is also hard to measure absorption rate constant is also hard to measure in patients, it is also desirable to avoid drawing in patients, it is also desirable to avoid drawing drug serum concentrations during the drug serum concentrations during the absorption phase in clinical situations. absorption phase in clinical situations.

3838

When only postabsorption, postdistribution serum When only postabsorption, postdistribution serum concentrations are obtained for a drug that is concentrations are obtained for a drug that is administered extravascularly, the equation simplifies administered extravascularly, the equation simplifies to: C = [(FD)/V]eto: C = [(FD)/V]e−ket−ket, where C is the concentration at , where C is the concentration at any postabsorption, postdistribution time; F is the any postabsorption, postdistribution time; F is the bioavailability fraction; D is the dose; V is the bioavailability fraction; D is the dose; V is the volume of distribution; kvolume of distribution; kee is the elimination rate is the elimination rate constant; and t is any postabsorption, postdistribution constant; and t is any postabsorption, postdistribution time. This approach works very well when the time. This approach works very well when the extravascular dose is rapidly absorbed and not a extravascular dose is rapidly absorbed and not a sustained- or extended-release dosage form. sustained- or extended-release dosage form.

3939

An example would be a patient receiving 24 An example would be a patient receiving 24 mEq of lithium ion as lithium carbonate mEq of lithium ion as lithium carbonate capsules. From previous clinic visits, it is capsules. From previous clinic visits, it is known that the patient has a volume of known that the patient has a volume of distribution of 60 L and an elimination rate distribution of 60 L and an elimination rate constant equal to 0.058 hconstant equal to 0.058 h−1−1. The bioavailability . The bioavailability of the capsule is known to be 0.90. The serum of the capsule is known to be 0.90. The serum lithium concentration 12 hours after a single lithium concentration 12 hours after a single dose would be: C = [(FD)/V]edose would be: C = [(FD)/V]e−ket−ket = [(0.90 24 ⋅ = [(0.90 24 ⋅mEq)/60 L] emEq)/60 L] e−(0.058 h−1)(12 h)−(0.058 h−1)(12 h) = 0.18 mEq/L. = 0.18 mEq/L.

4040

For example, a patient is given an oral dose of For example, a patient is given an oral dose of valproic acid 750 mg as capsules. Six and twenty-valproic acid 750 mg as capsules. Six and twenty-four hours after the dose, the valproic acid serum four hours after the dose, the valproic acid serum concentrations are 51.9 mg/L and 21.3 mg/L, concentrations are 51.9 mg/L and 21.3 mg/L, respectively. After graphing the serum respectively. After graphing the serum concentration/time data on semilogarithmic axes, the concentration/time data on semilogarithmic axes, the time it takes for serum concentrations to decrease by time it takes for serum concentrations to decrease by one-half can be measured and equals 14 hours. The one-half can be measured and equals 14 hours. The elimination rate constant is calculated using the elimination rate constant is calculated using the following equation: k= 0.693 / tfollowing equation: k= 0.693 / t1/21/2 = 0.693/14 h = = 0.693/14 h = 0.0495 h0.0495 h−1−1..

4141

Pharmacokinetic constants can also be calculated and used in Pharmacokinetic constants can also be calculated and used in these equations. If two or more postabsorption, postdistribution these equations. If two or more postabsorption, postdistribution serum concentrations are obtained after an extravascular dose, the serum concentrations are obtained after an extravascular dose, the volume of distribution, elimination rate constant, and half-life can volume of distribution, elimination rate constant, and half-life can

be computedbe computed

4242

The concentration/time line can be extrapolated to the The concentration/time line can be extrapolated to the y-axis where time = 0. Since this was the first dose of y-axis where time = 0. Since this was the first dose of valproic acid, the extrapolated concentration at t = 0 valproic acid, the extrapolated concentration at t = 0 (C(C00 = 70 mg/L) is used to estimate the hybrid volume = 70 mg/L) is used to estimate the hybrid volume

of distribution/bioavailability (V/F) parameter: V/F = of distribution/bioavailability (V/F) parameter: V/F = D/CD/C00 = 750 mg/70 L = 10.7 L. Even though the = 750 mg/70 L = 10.7 L. Even though the

absolute volume of distribution and bioavailability absolute volume of distribution and bioavailability cannot be computed without the administration of cannot be computed without the administration of intravenous drug, the hybrid constant can be used in intravenous drug, the hybrid constant can be used in extravascular equations in place of V/F. extravascular equations in place of V/F.

4343

An alternative approach is to directly calculate the parameters An alternative approach is to directly calculate the parameters without plotting the concentrations. The elimination rate without plotting the concentrations. The elimination rate constant (kconstant (kee) is computed using the following relationship: k) is computed using the following relationship: kee = = −(ln C−(ln C11 −ln C −ln C22)/(t)/(t11 −t −t22), where C1 is the first concentration at ), where C1 is the first concentration at time = ttime = t11, and C, and C22 is the second concentration at time =t is the second concentration at time =t22; k; kee = − = −[ln (51.9 mg/L) −ln (21.3 mg/L)] / (6 h −24 h) =0.0495 h[ln (51.9 mg/L) −ln (21.3 mg/L)] / (6 h −24 h) =0.0495 h−1−1. . The elimination rate constant can be translated into the half-The elimination rate constant can be translated into the half-life using the following equation: tlife using the following equation: t1/21/2 =0.693/k=0.693/0.0495 =0.693/k=0.693/0.0495 hh−1−1 =14 h. =14 h.

In this situation, the time/concentration pair at 24 hours will be In this situation, the time/concentration pair at 24 hours will be used (time =24 hours, concentration = 21.3 mg/L): Cused (time =24 hours, concentration = 21.3 mg/L): C00 =C/e =C/e−ket −ket

= (21.3 mg/L) / e= (21.3 mg/L) / e−(0.0495 h−1)(24 h)−(0.0495 h−1)(24 h) =70 mg/L. The hybrid volume of =70 mg/L. The hybrid volume of distribution/bioavailability constant (V/F) is then computed: distribution/bioavailability constant (V/F) is then computed: V/F =D/CV/F =D/C00 =750 mg/ (70 mg/L) =10.7 L. =750 mg/ (70 mg/L) =10.7 L.

4444

The hybrid constant volume of The hybrid constant volume of distribution/bioavailability (V/F) is computed by distribution/bioavailability (V/F) is computed by taking the quotient of the dose and the extrapolated taking the quotient of the dose and the extrapolated serum concentration at time = 0. The extrapolated serum concentration at time = 0. The extrapolated serum concentration at time = zero (Cserum concentration at time = zero (C00) is calculated ) is calculated using a variation of the intravenous bolus equation: using a variation of the intravenous bolus equation: CC0 0 = C/e= C/e−ket−ket, where t and C are a time/concentration , where t and C are a time/concentration pair that occur after administration of the pair that occur after administration of the extravascular dose in the postabsorption and extravascular dose in the postabsorption and postdistribution phases. Either valproic acid postdistribution phases. Either valproic acid concentration can be used to compute Cconcentration can be used to compute C00. .

4545

Calculation of Ka

1. Method of Residuals (good when Ka is at least 5 times Ke

•Starting with the equation for Cp versus time

Cp versus Time after Oral Administration this can be written as

Simplified Equation for Cp versus Time

4646

Semi-log plot of Cp versus Time after Oral Semi-log plot of Cp versus Time after Oral

AdministrationAdministration

4747

Semi-log Plot of Cp versus Time Showing CSemi-log Plot of Cp versus Time Showing Cpplatelate, Slope, and , Slope, and

InterceptIntercept

4848

and plotting Cplate versus time gives a straight line on

semi-log graph paper, with a slope (ln) = -kel and intercept = A. Now looking at the equation for Cp versus time again.

Difference or Residual versus Time

4949

5050

An Example Calculation Using the Method of ResidualsExample Data for the Method of Residuals

Time(hr)

PlasmaConcentration

(mg/L)

Cp(late)(mg/L)

Residual[Col3 - Col2]

(mg/L)

0.251.915.233.32

0.52.984.982.00

0.753.544.731.19

1.03.804.500.70

1.53.844.070.23

2.03.623.690.07

3.03.04

4.02.49

5.02.04

6.01.67

7.01.37

5151

5252

2. Wagner-Nelson Method2. Wagner-Nelson Method Advantages:Advantages: i) The absorption and elimination processes can be quite similar i) The absorption and elimination processes can be quite similar

and still accurate determinations of ka can be made. and still accurate determinations of ka can be made. ii) The absorption process doesn't have to be first order. This ii) The absorption process doesn't have to be first order. This

method can be used to investigate the absorption process.method can be used to investigate the absorption process.

Disadvantages:Disadvantages: The major disadvantage of this method is that you need to know The major disadvantage of this method is that you need to know

the elimination rate constant, from data collected following the elimination rate constant, from data collected following intravenous administration. intravenous administration.

5353

Amount absorbed = amount in body + amount eliminatedAmount absorbed = amount in body + amount eliminated

A = X + UA = X + U

Differentiation gives:Differentiation gives:

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Change in amount of drug in the body with time Change in amount of drug in the body with time is given by:is given by:

dX/dt = V.dCdX/dt = V.dCpp/dt/dt Change in amount of drug eliminated is given Change in amount of drug eliminated is given

by:by: dU/dt = V.kdU/dt = V.kelel.C.Cpp

Therefore, dA/dt = V.dCTherefore, dA/dt = V.dCpp/dt + V.k/dt + V.kelel.C.Cpp

dA = V.dCdA = V.dCpp + V.k + V.kelel.C.Cp p dtdt

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Where, A is the amount absorbed up to time tWhere, A is the amount absorbed up to time t Taking this to infinity gives the maximum amount Taking this to infinity gives the maximum amount

absorbed (since Cabsorbed (since Cpp equals zero after a time equals equals zero after a time equals

infinity:infinity:

AAmaxmax = 0 + V . k = 0 + V . kelel . AUC . AUC0-0-

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The amount remaining to be absorbed (AThe amount remaining to be absorbed (Amaxmax – A), can – A), can

also be expressed as the amount remaining in the GI, also be expressed as the amount remaining in the GI, XXgg

XXgg = (A = (Amaxmax – A), – A), Dividing by V we get:Dividing by V we get: XXgg/V = {(A/V = {(Amaxmax/V) –(A/V)}/V) –(A/V)} If the kinetics governing the process is a 1If the kinetics governing the process is a 1stst order one, order one,

then:then: (X(Xgg/V) = (X/V) = (Xg0g0/V) e/V) e-kat-kat X/V = ConcentrationX/V = Concentration

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It follows that:It follows that: Plotting (APlotting (Amaxmax-A)/V versus time produces a -A)/V versus time produces a

straight line on semi-log graph paper and a straight line on semi-log graph paper and a curved line on linear graph paper. From the curved line on linear graph paper. From the slope of the line on the semi-log graph paper slope of the line on the semi-log graph paper kkaa can be calculated (slope = K can be calculated (slope = Kaa) )

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CCpopo = k = kelel . AUC . AUC0-0-= A= Amaxmax

CCp,tp,t = k = kelel . AUC . AUC0-t0-t

AUCAUCt1-t2t1-t2 = {(C = {(Cp1p1+C+Cp2p2)/2}(t)/2}(t22 – t – t11) = Column 3) = Column 3 Total AUC = Column 4Total AUC = Column 4

A/V = CA/V = Cp p + K+ Kelel AUC AUC0-t0-t

CCpp = column 2 = column 2

KKelel AUC AUC0-t0-t = Column 5 = Column 5

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