DEVELOPMENT AND EVALUATION OF AZITHROMYCIN DIHYDRATE FAST DISSOLVING TABLET Jyotsana Madan1, Pradip Kumar Karar2, Gaurav Agarwal*2 1 SKN College of Pharmacy, Pune (Maharastra) India 2 Faculty of Pharmacy, RPIIT Campus, Karnal (Haryana) India ABSTRACT The aim is to formulate fast dissolving tablets of Azithromycin Dihydrate which is intended to disintegrate rapidly into the oral cavity using different super disintegrants. The wet granulation method was used because direct compression method failed to formulate dispersible tablet. The different super disintegrant used were Sodium starch glycolate (SSG), Croscarmellose sodium (CCS) and Crospovidone (CPV). In all the formulations water was used as a binding agent. The finally prepared fast …show more content…
The Prepared fast dissolving tablets should disintegrate rapidly in oral cavity within 15-60 seconds, without water. The super disintegrant ( Sodium starch glycolate, Croscarmellose sodium & Crospovidone) were used to formulate the tablets. Fast dissolving tablet of Azithromycin dihydrate were prepared by wet granulation method according to the formula given in Table 1. Initially all the ingredients were passed through 66 mesh separately. Then the granules were prepared by using SSG,CCS & CPV, sodium lauryl sulphate and avicel as intragranular ingredients using water as a binder passing lumps through 8 mesh. Then extragranular ingredients were weighed and mixed in geometrical order with prepared granules and compressed into tablet of 400 mg using 11mm flat punches on single punch tablet compression machine. A batch of 30 tablets were prepared for each of the designed …show more content…
Preformulation yields basic knowledge necessary to develop suitable formulation for the toxicological use. It gives information needed to define the nature of the drug substance and provide frame work for the drug combination with pharmaceutical excipients in the dosage form. The following preformulation studies were performed. 1. Bulk Density (Db): It is the ratio of total mass of powder to the bulk volume of powder. It was measured by pouring the weight powder (passed through standard sieve # 20) into a measuring cylinder and initial weight was noted. This initial volume is called the bulk volume. From this the bulk density is calculated according to the formula mentioned below. It is expressed in g/ml and is given by Db = M/ Vb Where, M is the mass of powder, Vb is the bulk volume of the powder. 2. Tapped Density (Dt): It is the ratio of total mass of the powder to the tapped volume of the powder. Volume was measured by tapping the powder for 750 times and the tapped volume was noted if the difference between these two volumes is less than 2%. If it is more than 2%, tapping is continued for 1250 times and tapped volume was noted. Tapping was continued until the difference between successive volumes is less than 2 % (in a bulk density apparatus). It is expressed in g/ml and is given
It is soluble in water and N,N-dimethyl formamide; slightly soluble in methanol; very slightly soluble in ethanol, acetone, and acetonitrile; and insoluble in isopropanol and isopropyl
Prelab week 1 Calculations Preparation of 1.5μmol/L mixed low-level standard dilution 150μmol/L × V1=1.5μmol/L × 10ml V1=(1.5μmol/L×10ml)/(150μmol/L)=0.1ml Conversion of milliliters to microliters (0.1ml×1000)μL= 100μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=3μmol/L × 10ml V1=(3μmol/L×10ml)/(150μmol/L)=0.2ml Conversion of milliliters to microliters (0.2ml×1000)μL= 200μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=7.5μmol/L × 10ml V1=(7.5μmol/L×10ml)/(150μmol/L)=0.5ml Conversion of milliliters to microliters (0.5ml×1000)μL= 500μL Preparation of the blank samples The volumetric flask will be filled to the mark with 150μmole/L of stock solution to act as blank (reference). Additional two blanks will
Marwah Alabbad Post lab 10/21/15 Question 1: 1. Experiment 1: Number of trails NaOH concentration (M) Volume of HCl solution (mL) Initial volume of NaOH(mL) final volume of NaOH(mL) The volume of NaOH to titrate HCl (mL) Concentration of HCl (M) 1st 0.1023 25.0 10.05 36.12 26.07 0.085 2nd 0.1023 25.0 5.74 31.40 25.66 0.105 3rd 0.1023 25.0 9.84 35.52 25.68 0.105 First trail calculation: 0.02607L× (0.1023mole NaOH/1L)×(1 mol of HCL/1 mol of NaOH)×(1/0.025)= 0.085M of HCl
Fill beaker with water Use the disposable pipette to place water in the graduated cylinder until the unidentified object would be completely submerged in water Record what the measurement of water in milliliters before placing the unidentified object into the graduated cylinder Gently place the unidentified object into the graduated cylinder Record the measurement of the water in milliliters after placing the unidentified object into the graduated cylinder Subtract the measurement of water in milliliters before placing the unidentified object into the graduated cylinder from the measurement of the water in milliliters after placing the unidentified object into the graduated cylinder, this is the volume of the unidentified object Record the volume (the answer you got in step 10) of the unidentified object in the data table Weigh the unidentified object on the scale, this is the mass of the unidentified object Record that number in the data table Calculate the density of the object by dividing the mass by the volume and rounding it to the proper significant figure, Record the density of the unidentified object in the data table Repeat the lab 2 more times and with each experiment record the data in the chart under the correct trial number corresponding with the correct
By reading the new volume of the liquid substance amount one will then subtract the initial milliliter amount from the final volume reading, thus giving you the volume of the rock sample. Using the mass of the sample rock obtained one will then divide the final volume reading unveiling the density of the
Therefore, liquid-liquid and acid-base extraction techniques were successfully performed to separate the components of the Excedrin tablet. According to the TLC analysis results, the compounds (aspirin, acetaminophen, and caffeine) were successfully isolated from the analgesic (Excedrin tablet). In figure 1, the separation of the compound in the TLC analysis correlates with the TLC analysis in figure 2. Furthermore, Rf index calculations of the TLC analysis demonstrated that the compounds (aspirin, acetaminophen, and caffeine) were separated. The Rf calculations of aspirin in table 1 shows an Rf value of .491; however, in table 2 the Rf value of aspirin was calculated to be .784.
In addition, for calculating the mass of the pennies and the unknown substance, zero the balance and place it in the weigh boat to receive the data. Lastly, to calculate the density of the substances, use the formula D=MV, in which dividing the mass by the volume allows to do so. In order to be certain of the data that is collected, running multiple trials could help be accurate. A method to get an average value of the density can be expressed by the formula D1+D22. This formula basically allows one to add the data of all the trials and divide it by the number of trials that was performed.
As much was conducted throughout this lab, the projected completion of this lab displays that ultimately, the higher the temperature of the water, the faster the dissolving rate of the Alka-Seltzer is. In other words, the hotter the water temperature the quicker the tablet dissolves within the water in regards to the amount of time it took to dissolve. Furthermore, this experiment helps to explain that, if water is taken at a higher temperature and Alka-Seltzer is placed within the water, the Alka-Seltzer will take less time to dissolve because the higher temperatures cause the tablet to melt at a quicker rate. This compares to when Alka-Seltzer is placed in colder temperatures, where instead it takes more time to dissolve, because the lower
highlighted that in United States, the dose dumping in general and alcohol induced dose dumping in particular is considered as a serious concern for orally administered prolonged release dosage forms (3). Subjuct to the therapeutic
To calculate the experimental mass the substance of each bag and the bag its self was measured using a balance. After gathering the mass subtract the mass of the empty bag to the mass of the unknown substance, in order to just have the mass of the substance. Afterward the mass of the unknown substance was divided by the number of moles recorded on the bag of the substance. The measurements are displayed on the table
Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data
75.33 grams Weight of the unknown = 0.23 grams Calculation : 75.33-75.10/0.36x100 = 63.8 % recovery Melting point of
Acids are proton donors in chemical reactions which increase the number of hydrogen ions in a solution while bases are proton acceptors in reactions which reduce the number of hydrogen ions in a solution. Therefore, an acidic solution has more hydrogen ions than a basic solution; and basic solution has more hydroxide ions than an acidic solution. Acid substances taste sour. They have a pH lower than 7 and turns blue litmus paper into red. Meanwhile, bases are slippery and taste bitter.
Introduction Buffer is a solution that resists a change in pH when bases or acid are added. Solutions that are acidic contain high concentrations of hydrogen ions (H+) and have pH values less than seven. Buffer usually consist of a weak acid, and its conjugate base or a weak base and its conjugate acid. The function of buffer is to resist the changes in hydrogen ion concentration as a result of internal and environmental factor. This buffer experiment is important so that we relies the important of buffer in our life.