Injectable Drugs and Their Impact – Drug Therapy Example

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"Injectable Drugs and Their Impact" is a perfect example of a paper on drug therapy.   A great number of injectable drugs fail to dilute when mixed with  infusions or  in  syringes. Infusion of other injectable drugs in infusion bags is quite unsafe. At times, their incompatibility involves the formation of precipitates, the evolution of gases, reactions of ions, and biological molecules'  denaturation. It is important to know which drugs. Inspections of incompatibilities are vital despite some of the incompatibilities being invisible. This analysis helps to decide what to undertake in such situations.     Henderson-Hasselbalch equation  Phenytoin is a weak acid  pKa  = 8.1  IV infusion  pH    =  5.0  pH =  pKa  + log ([A-]/[HA])    5.0 = 8.1 + + log ([A-]/[HA])  -3.1= log ([A-]/[HA])  0.0008 = ([A-]/[HA])  The unionized phenytoin is not soluble in water, hence forming a precipitate in the weak acid environment.   The  concentration of phenytoin in the IV solution is in molar terms.

It is therefore wise to increase the volume of the IV solution to increase solubility in the two drugs. This is however not realistic in practical situations because increasing the dosage of parenteral solutions may cause harm to the patient.     [A-] = molar concentration of a conjugate base    [HA] = molar concentration of an  undissociated  weak acid (M)    Phenytoin is a weak acid mixing with a base.

Hence, the pharmacist must administer it gradually at a rate of 50mg in every minute.           Gradient= (change in y)  /(change in x)  =(6.7-4.5)/(4-6)  =2.2/-2  =-1.1  At  point 0 (x, 0);   -1.1=(0-2)/(x-10)  -1.1x+10=-2  -1.1x=-12  1.1x=12  X=10.9090909  Cp(0)=10.909090  Calculating the volume of the drug:   t1/2 = (0.693 x  Vd)  /CL  6 = (0.693 x  Vd)  / (60mg/hr)  360 hours = 0.693 x  Vd  Vd= 519.4805 mg  The  effect of a  chemical structure of a compound is direct and affects the chemical, physical and biological properties of a patient. One of the processes and concentrations of body fluids it alters is the plasma concentration in the blood.

Intervals of induction of drug X results in the diminishing of the volume of plasma in the blood.     Vd  is  important  for a prescriber because it enables the prescriber  to know  and understand how long a drug stays in a patient and what capacity or amount needs to be added. Some drugs take longer to diminish because they are fat-soluble while others take much less time because they are water-soluble. Too much drug in a patient may lead to  the development  of complications and possibly death.   Elimination rate constant =  Kel  = CL /  Vd  = (30/ 15 mg/L/ 519.4805 mg    = 0.00385 mg/L  Calculating the clearance rate of drug X  CL=  Cl = (Dose / interval) /  Cpss  ave  = (60mg/2hrs) / 15mg/L  = 30/15  =2 mg/hr  Half life  is the time which an administered drug takes to reduce in its concentration by half.   T ½ = 0.693 /  Kel  = 0.693/0.00385  = 180hr  Clearance is a measure of how the renal excretes water and acidic matter from our body after its use.

Different substances have different filtration rates which determine the rate of excretion of that  particular substance.    


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