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B C D A B C D A FUNCTIONSe.g. TEMPERATURE;TEMPERATURE; HEART RATE;HEART RATE; BLOOD PRESSURE.BLOOD PRESSURE.

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Presentation on theme: "B C D A B C D A FUNCTIONSe.g. TEMPERATURE;TEMPERATURE; HEART RATE;HEART RATE; BLOOD PRESSURE.BLOOD PRESSURE."— Presentation transcript:

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2 B C D A

3 B C D A

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7 FUNCTIONSe.g. TEMPERATURE;TEMPERATURE; HEART RATE;HEART RATE; BLOOD PRESSURE.BLOOD PRESSURE.

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9 A B C

10 Isotonic Solutions contain the same concentration of solute as an another solution (e.g. the cell's cytoplasm). When a cell is placed in an isotonic solution, the water diffuses into and out of the cell at the same rate. The fluid that surrounds the body cells is isotonic.

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23 The Osmosis definition : Osmosis is the passage of water from a region of high water concentration through a semi-permeable membrane to a region of low water concentration.

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30 FACT THE OSMOTIC PRESSURE OF OUR BODY FLUIDS, EQUIVALENT TO NaCl SOLUTION. ?

31 WHY NaCl?WHY NaCl? WHY ONLY 0.9%?WHY ONLY 0.9%? Where this (0.9%) came from?Where this (0.9%) came from?

32 Na &Cl are the most plentiful electrolyte in the body.Na &Cl are the most plentiful electrolyte in the body. NORMAL HEALTHY HUMAN: Na mMol.L -1NORMAL HEALTHY HUMAN: Na mMol.L -1 Cl mMol.L -1 K mMol.L -1 Ca mMol.L -1 Cl mMol.L -1 K mMol.L -1 Ca mMol.L -1

33 WHY ONLY 0.9%? FREEZING POINT o C

34 WHY ONLY 0.9%? FREEZING POINT o C

35 Mole in gL -1.= o C*i XgL o C For NaCl 58.5gL -1 = o C*1.8 XgL o C X= g L -1.

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37 Count Lorenzo Romano Amedeo Carlo Avogadro diQuaregna e Cerreto, (TurinAugust 9, July 9, 1856)TurinAugust 91776July u “Volumiegualidi gas nellestessecondizionidi temperatura e dipressionecontengono la stessonumerodimilecol e” Amedeo Avogadro

38 AMEDEO AVOGADRO ( ) u “Equal volumes of gases at the same temperature and pressure contain the same number of molecules” Amedeo Avogadro

39 AVOGADRO’S NUMBER 6.02x10 23

40 AVOGADRO’S NUMBER THE NUMBER OF PARTICLES IN A SOLUTION OF ONE Kg OF WATER.

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43 Osmol THE WEIGHT IN g. OF A SOLUTE. EQUIVALENT TO A MOLE.

44 mOsm THE WEIGHT IN mg OF A SOLUTE. EQUIVALENT TO A mMOLE.

45 Osmol IT IS THE AMOUNT OF A SOLUTE, WHICH WILL PROVIDE ONE AVOGADRO’S NUMBER

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49 Nonelectrolyte(1) Boric acid Dextrose Glycerin Mannitol monks or nuns

50 Mole in gL -1.= o C*i XgL o C For Boric acid 61.8gL -1 = o C*1 XgL o C X= g L -1.

51 EQUIVALENT TO

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53 WHY? NONELECTROLYTE

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55 FREEZING POINT “ D ” EPRESSION i = Dissociation(80%) NaClNa + Cl 20%80% + 80% 180/100 = 1.8

56 NUMBER of PARTICLES

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59 FREEZING POINT “ D ” EPRESSION i = Dissociation(80%) KCl K + Cl 20%80% + 80% 180/100 = 1.8

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61 FREEZING POINT “ D ” EPRESSION i = Dissociation(40%) ZnSO 4 Zn + SO 4 60%40% + 40% 140/100 = 1.4

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63 FREEZING POINT “ D ” EPRESSION i = Dissociation CaCl 2 Ca +2 Cl 20%80% + (2*80%) 260/100 = 2.6

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65 FREEZING POINT “ D ” EPRESSION i = Dissociation K 2 SO 4 2 K+SO 4 20%(2*80%) + 80% 260/100 = 2.6

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67 FREEZING POINT “ D ” EPRESSION i = Dissociation FeCl 3 Fe+3CL 20%80% + (3*80%) 340/100 = 3.4

68 ISO-OSMOTIC or ISOTONIC or

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70 Why?

71 ISOTONICHYPERTONIC

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75 gL -1 BORIC ACID

76 Boric Acid Pass Freely Through the RBC MembraneRegardless of Concentration. WHY?

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79 LAXATIVE MAGNESIUM SULFATEMAGNESIUM SULFATE MAGNESIUM CITRATEMAGNESIUM CITRATE GLYCERIN [RECTAL]GLYCERIN [RECTAL]

80 LACTULOSE & CEPHULAC NON-ELECTROLYTENON-ELECTROLYTE NONABSORBABLE DISACCHARIDE, NONABSORBABLE DISACCHARIDE, + COLON BACTERIA LACTIC ACID+ COLON BACTERIA LACTIC ACID OSMOTIC PRESSUREOSMOTIC PRESSURE ACIDIFICATION SERVE AS A TRAP FOR AMMONIA BLOOD LEVELACIDIFICATION SERVE AS A TRAP FOR AMMONIA BLOOD LEVEL Rx: SYSTEMIC ENCEPHALOPATHYRx: SYSTEMIC ENCEPHALOPATHY Class Generic Name Brand Name Ammonia DetoxicantsLactuloseCEPHULAC

81 S&S of  OSMOTICITY Normal 285 mOsmol kg mOsmol kg -1 THIRSTY mOsmol kg -1 DRY MUCOUS MEMBRANE: mOsmol kg -1 WEAKNESS, DOUGHY SKIN: mOsmol kg -1

82 >330 mOsmol kg -1 DISORIENTATIONDISORIENTATION POSTURAL HYPOTENSIONPOSTURAL HYPOTENSION SEVERE WEAKNESSSEVERE WEAKNESS FAINTINGFAINTING COMACOMA

83 if But what O.P. 1%

84 HEADACH : mOsmol kg -1 DROWSINESS: mOsmol kg -1 Disorientation CRAMPS mOsmol kg -1 <230 mOsmol kg -1 SEIZURES & COMA

85 SERUM OSMOLALITY 1.86 Na + BLOOD SUGAR + BUN Na + BLOOD SUGAR + BUN 20 3 A QUIKY 2 Na + 10 THE QUIKIST

86 WHOLE MILK 295 TOMATO JUICE 595 ORANGE JUICE935

87 PITUITARY ANTIDIURETIC HORMONE (ADH) SERUM OSMOCITY

88 OSMOLALITY D ETERMINATION

89 OS NaCl “E”QUIVALENT  in RBC FREEZING POINT

90 OS

91 FREEZING POINT “ D ” EPRESSION o C NaCl 0.9%

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93 Rx Naphazoline HCl0.02% [ 247,2] Zinc sulfate0.25% Purified waterqs30 mL Mft Isotonic solution

94 FREEZING POINT D

95 FREEZING POINT “ D ” EPRESSION Mole in gL -1 = o C*i XgL -1 - o C Naphazoline HClM.Wt.247 Ions2 Rx Naphazoline HCl0.02% [ 247,2] Zinc sulfate0.25% water qs30 mL Mft Isotonic solution

96 FREEZING POINT “ D ” EPRESSION 247gL -1 =-1.86 o C* gL -1 - X o C Naphazoline HCl o C

97 FREEZING POINT “ D ” EPRESSION 288gL -1 = o C* gL -1 - X o C ZnSO o C

98 Total depression o C o C o C

99 = ( ) = o C = = o C

100 270 mg0.52 o C Xmg o C

101 Step 1  Equation How much depression it can a drug cause? 2  of contribution drug  3

102 OS NaCl “E”QUIVALENT  in RBC FREEZING POINT

103 The weight of NaCl which will produce the same osmotic pressure effect as 1 g. of the drug.

104 Rx Naphazoline HCl0.02% Zinc sulfate0.25% Purified waterqs30 mL Mft Isotonic solution

105 Mole of NaCl gL -1 * i Drug Mole of drug gL -1 * i NaCl Naphazoline HClM.Wt.247 Ions2

106 247gL -1 *1.8 “E” forNaphazoline HCl 58.5gL -1 *1.8 = g NaCl

107 Rx Naphazoline HCl 0.02 g* 30 mL/100 mL= g Naphaz g NaCl*0.006 g Naph= g NaCl 1 g Naphz Rx Naphazoline HCl 0.02% Zinc sulfate0.25% Purified waterqs 30 mL Mft Isotonic solution Rx Naphazoline HCl 0.02% Zinc sulfate0.25% Purified waterqs 30 mL Mft Isotonic solution

108 288gL -1 *1.8 “E” forZnSO gL -1 *1.4 =

109 RxZnSO *30/100=0.075 g Zn SO *0.075= g NaCl Rx Naphazoline HCl 0.02% Zinc sulfate0.25% Purified waterqs 30 mL Mft Isotonic solution Rx Naphazoline HCl 0.02% Zinc sulfate0.25% Purified waterqs 30 mL Mft Isotonic solution

110 0.075 g Zn SO 4 = g NaCl g Naphaz.HCl = g NaCl =0.0132g NaCl 

111 9 mgmL -1 NaCl*30 mL=270 mg NaCl. If there no medication, this prescription will be isotonic with 270 mg of NaCl.

112 Rx NaCl270 mg Purified waterqs30 mL Mft Isotonic solution Rx NaCl270 mg Purified waterqs30 mL Mft Isotonic solution Samaan

113 NaCl270 mg NaCl 13.2 mg NaCl mg

114 256.8 mg NaCl with “E” 257 mg NaCl with “D”

115 Rx Naphazoline HCl6 mg Zinc sulfate75 mg Sodium chloride257 mg Purified waterqs30 mL Rx Naphazoline HCl6 mg Zinc sulfate75 mg Sodium chloride257 mg Purified waterqs30 mL Samaan

116 OS NaCl “E”QUIVALENT  in RBC FREEZING POINT

117 V olumeof water to be added to a specified weight of drug to prepare an isotonic solution.

118 Rx Naphazoline HCl0.02% Zinc sulfate0.25% Purified waterqs30 mL Mft Isotonic solution Rx Naphazoline HCl0.02% Zinc sulfate0.25% Purified waterqs30 mL Mft Isotonic solution

119 V

120 Rx Naphazoline HCl0.02% Rx Naphazoline HCl0.02%

121 247gL -1 *1.8 “E” for Naphazoline HCl 58.5gL -1 *1.8 = g NaCl = g NaCl* 0.006g naph.HCl= g NaCl

122 0.006 g N.HCl= g NaCl mL Dissolve 6 mg of Naph.HCl in 0.16 of water, the sol. will be isotonic g NaCl* 1 ml= mL H 2 O gNaCL

123 Rx Zinc sulfate0.25% Rx Zinc sulfate0.25%

124 288gL -1 *1.8 “E” for ZnSO gL -1 *1.4 = g NaCl = g NaCl* g = g NaCl* g ZnSO 4 = g NaCl

125 1.31 mL 1.31 mL If you dissolve 75 mg of ZnSO 4 in 1.31 mL of H 2 O, the sol. will be isotonic g NaCl/0.009 g NaCl=

126 mL mL mL

127 Dissolve 6 mg of naph.HCl& 75 mg ZnSO 4 in 1.47 mL water, and qs to 30 mL with isotonic 0.9% NaCl.

128 PROBLEM! WHAT PROBLEM?

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130 P1 BORIC ACID “E” 0.52 Isotonic NaCl solution contains 0.9% w/v NaCl. If the “E” value of boric acid is 0.52, calculate the % strength (w/v) of an isotonic solution of boric acid.

131 BORIC ACID “E” g. of boric acid has the same osmotic pressure as 0.52 g. of NaCl.

132 IF 0.9% NaCl IS ISOTONIC 1 g OF BORIC ACID = 0.52 g NaCl X g OF BORIC ACID= 0.9 g NaCl X g BORIC ACID = 1 g B.A. * 0.9 g 0.52 g = 1.73 g%

133 Mole gL -1 = o C*i x gL o C 61.8gL -1 = o C*1.0 x gL o C = 17.27g L -1

134 17.27 g 1000 mL x g 100 mL 1.727g%

135 P2 IF NaCl DISSOCIATING AT 90%. CALCULATE:- A- DISSOCIATION FACTOR; B- FREEZING POINT OF MOLAL SOLUTION.

136 i =Dissociation(2) NaClNa + Cl 10%90% + 90% 190/100 = 1.9 A-

137 - X o C = o C *1.9 B- X = o C

138 P3 WHAT IS THE F.P. OF 25 g IN 500 mL DEXTROSE? D5W

139 Mole gL -1 = o C*i [ gL -1 ] - X o C 180 gL -1 = o C*1 50 gL -1 - X o C = = o C

140 P4 PROCAINE HCl [M.Wt 273; 2-ION] DISSOCIATING AT 80%. A- DISSOCIATION FACTOR, B-”E” C- F.P. FOR A MOLAL SOLN.

141 A- DISSOCIATION FACTOR, PROCAINE HCl PROCAINE + HCl 20%80% + 80% 180/100 =1.8

142 B- “E” 273gL -1 * gL -1 *1.8 = g Procaine HCl= g NaCl

143 C- F.P. FOR A MOLAL SOLUTION = o C*i = o C *1.8 = = o C

144 P5 The freezing point of a molal solution of a nonelectrolyte is -1.86°C. What is the freezing point of a 0.1 % solution of zinc chloride (M.Wt. 136), dissociating 80%?

145 F.P. OF 0.1%ZnCl 2 [MWt=136,3] Mole gL -1 = o C*i [gL -1 ] - X o C 136 gL -1 = o C*2.6 1gL -1 - x o C = = o C F.P.D=

146 P6P6P6P6 F.P. of 5% boric acid is o c. HOW many g. of boric acid should be used to prepare one L of an isotonic sol.?

147 Mole gL -1 = o C*i XgL o C 50 gL -1 = o C*1 XgL o C gL -1

148 Mole gL -1 = o C*i XgL o C 6I.8 gL -1 = o C*1 XgL o C gL -1

149 P7Rx Ephedrine sulfate 300 mg Sodium Chlorideq.s. Purified waterad 30 mL Make isotn.sol. Sig. Use as directed How many mg. of NaCl? Eph.SO 4 429,3 [429,3]

150 1 R How much NaCl can make the whole Rx isotonic?  Rx of contribution drug 3Step 2  R-Rx

151 429gL -1 * gL -1 * 2.6 = g Nacl NaCl ‘E’ 0.3 g * = g NaCl= 59 mg NaCl 270 mg - 59 = mg NaCl 1 Step  2  9 mg NaCl* 30 mL the total volume of the Rx 1mL = 270 mg of NaCl R Rx Ephedrine sulfate has the same osmotic pressure as g Nacl 1 g of Ephedrine sulfate has the same osmotic pressure as g Nacl  3

152 P8P8P8P8 Rx Dipivefrin HCl0.5% [388,2] Scopolamine HBr0.33[438,2] SodiumChlorideq.s. Purified waterad 30.0 mL Make isotn.sol. Sig. Use in the eyes. How many g. of NaCl?

153 388gL -1 * gL -1 *1.8 = 0.15 gNaCl 0.5 g100 mL x g. 30 mL 0.15 g * 0.15 = g Dipivefrin HCl 22.5 mgNaCl E

154 Scopolamine HBr 438gL -1 * gL -1 *1.8 = gNaCl 0.33 g100 mL x g. 30 mL 0.1 g * = gNaCl E

155 g g g Both drugs ‘E’

156 0.270 g g g NaCl reference-‘E’

157 P9P9P9P9 RxZinc sulfate 0.06 Boric acidq.s. Purified waterad 30mL Make isotn. sol. Sig. drop in eyes How many g. of boric acid?

158 “E” for Zinc sulfate 288gL -1 * gL -1 *1.4 = * = g NaCl

159 0.270 g g g NaCl reference - ‘E’ but Rx calls for boric acid!

160 1 g Boric acid0.52 g NaCl x g Boric acid g NaCl g Boric acid

161 “EB” for Zinc sulfate 288gL -1 * gL -1 *1.4 = 0.3 g B.A. 0.06* 0.3= g Boric acid 17.3 mg*30 mL 1 mL Boric acid isotonic reference =519 mg of Boric acid to make 30mL isotonic [reference]

162 0.519 g of B.A. [reference] g of boric acid equivalent g Boric acid

163 P RxCromolyn Na4% [512,2] Benzalkonium Cl. (1:10,000)[360,2] Buffer solq.s. Purified waterad 10mL Make isotn. sol. Sig. One drop in each eye How many mL of the buffer solution (E = 0.30) should be used to render the solution isotonic?

164 “E” for Cromolyn Na 512gL -1 * gL -1 *1.8 = g NaCl 0.4 g C.Na* = g NaCl R= 9 mg NaCl* 10 mL/1 mL = 90 mg NaCl

165 “E” for Benzalkonium Cl. (1:10,000)[360,2] 360gL -1 * gL -1 *1.8 = g NaCl 0.001g Bz.* = g NaCl 0.09 g NaCl g NaCl g NaCl

166 (E = 0.30) u 1 g of buffer material = 0.3 g of NaCl  g NaCl *  g NaCl * 1 g of buffer material 0.3 g of NaCl = g of buffer material 1 g of buffer material = 0.3 g of NaCl 3 g of buffer material = 0.9 g of NaCl 3 % of buffer sol. = 0.9 % of NS

167 4.9 mL of isotonic buffer solution g of buffer X mL g Buffer solution 100 mL X mL of isotonic buffer solution =0.147 g * 100 ml/3.0 g buffer= 4.9 mL

168 P11 RxDextrose,anhydrous 2.5% NaClq.s. Sterile water for injection ad 1000mL Label: isotonic Dextrose & Saline Solution. How many g of NaCl needed?

169 “E” for anhydrous Dextrose 180gL -1 * gL -1 *1 = 0.18 g NaCl 25 g.* 0.18= 4.51g NaCl 4.5 g NaCl

170 P Rx sol.Silver Nitrate 0.5%15.0 Make isoton. sol. Sig. For the eyes. How many g of KNO 3 needed? Why not to use NaCl as adjustor?

171 Reference of KNO 3 Mole gL -1 = o C*i XgL o C 101gL -1 = o C*1.8 XgL o C gL -1

172 Reference of KNO g 1000 mL x g15 mL g of KNO 3 to fill up this prescription without any medication.

173 “E KNO3 ” for Silver Nitrate 170gL -1 * gL -1 *1.8 =0.5941gKNO g AgNO 3 * = g KNO g of KNO 3 Reference g of KNO 3 Rx 0.190g of KNO 3

174 P13 Rx Cocaine HCl0.15[340,2] NaClq.s. Purified Water ad15 Make isoton. sol. Sig. One drop for the left eye. How many g of NaCl needed?

175 “E” for Cocaine HCl 340gL -1 * gL -1 *1.8 = 0.172g NaCl 0.15 g C.HCl* 0.172= g NaCl R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl R-Rx= g g= 0.109gNaCl

176 P14 Rx Cocaine HCl0.6 [340,2] Eucatropine HCl0.6 [328,2] Chlorobutanol0.1 [177,1] NaClqs Purified Waterad 30 Make isoton. sol. Sig. For the eyes.

177 58.5gL -1 * g CHCl.* 0.172= g NaCl “E” for Cocaine HCl 340gL -1 *1.8 = 0.172g NaCl R=0.009g NaCl * 30 mL/1 mL =0.27 g NaCl R=0.009g NaCl * 30 mL/1 mL =0.27 g NaCl

178 “E” for Eucatropine HCl 328gL -1 * gL -1 *1.8 = g Euc.* 0.178= g NaCl

179 “E” for 177gL -1 * gL -1 *1.0 = g Chl.* = g NaCl Chlorobutanol

180 Cocaine HCl g Eucatropine HCl g Chlorobutanol g g g g = g

181 P15 RxTetracaine HCl0.1 [301,2] Zinc sulfate0.05 [288,2] Boric acid qs Purified Waterad 30 Make isoton. sol. Sig. For the eyes. How many g. Boric acid needed?

182 “E b ” for Tetracaine HCl 301gL -1 * gL -1 *1.8 = g Boric 0.1 g Tetr.* = g Boric 17.3 g*30 mL= g Boric 1000 mL Reference

183 “E b ” for Zinc sulfate 288gL -1 * gL -1 *1.4 = g Boric 0.05 g Zn.* = g Boric

184 Tetracaine HCl Zinc sulfate g g g g Boric acid Equivalent g Boric acid Reference g Boric acid

185 P16 Rx Sol. HomatropineHBr 1% 15 [356,2] Boric acid q.s. Make isoton. sol. Sig. For the eyes. How many g. Boric acid needed?

186 “E b ” for HomatropineHBr 356gL -1 * gL -1 *1.8 = g*15 mL/1000 mL Boric acid Reference g Boric acid x g = 0.15 g

187 0.15 g Homa.* = g Boric Boric acid Equivalent g Boric acid reference g Boric acid Equivalent g Boric Acid

188 P17 Rx Procaine HCl 0.1% [ NaCl q.s. Procaine HCl 0.1% [273,2] NaCl q.s. SterileWater for Inj. ad100.0 Make isoton. sol. Sig. For Injection. How many g. NaCl needed?

189 “E” for 273gL -1 * gL -1 *1.8 = g NaCl 1 g Proc.* = g NaCl Procaine HCl R=0.009g NaCl * 100 mL/1 mL =0.9 g NaCl 0.9 gNaCl reference g NaCl Equival g NaCl

190 P18 Rx Phenylephrine HCl 1%[204,2] Chlorobutanol 0.5%[177,1] NaCl q.s. Purified Water ad 15.0 Make isoton. sol. Sig. Use as directed How many mL NSS needed?

191 “E” for Phenylephrine HCl 204gL -1 * gL -1 *1.8 = g NaCl 0.15 g Phen.* = g NaCl R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl

192 “E” for 177gL -1 * gL -1 *1.0 = g NaCl g Ch.* = Chlorobutanol g NaCl Phenylephrine HCl + Chlorobutanol g NaCl g NaCl = g NaCl g NaCl*100 mL 0.9 g NaCl R-Rx= =0.078 g NaCL 8.69 mL N.S.

193 P19 Rx Oxymetazoline HCl 0.5% [297,2] Boric acid sol. qs Purified Water ad 15.0 Make isoton. sol. Sig. For the nose, as decongestant How many mL of 5% boric acid solution needed?

194 “Eb” for 297gL -1 * gL -1 *1.8 = g B.A g Oxy* = g B.A. Oxymetazoline HCl 17.3 g*15 mL= g B.A mL R

195 gB.acid reference g B.acid Equival g Boric acid 4.63 mL of 5% Boric acid solution g B.A.*100 ml/5 g B.A.

196 P20 Rx Ephedrine HCl 0.55 [202,2] Chlorobutanol0.25 [177,1] Dextrose qs Rose Water ad 50.0 Make isoton. sol. Sig. Nose drop. How many g of Dextrose needed?

197 180 g L -1 = 1.86 *1 x g L o C gL g*0.05 L 1L g Dex. Ref.

198 “Ed” for Ephedrine HCl 202gL -1 * gL -1 *1.8 = g Eph* 1.6= 0.80g Dex.

199 “Ed” for Chlorobutanol 177gL -1 * gL -1 *1.0 = g Ch* = g Dex.

200 Ephedrine HCl 0.80g Dex. Chlorobutanol g Dex g Dex g Dex. Equival g Dex g Dex. Ref.

201 P21 Naphzoline HCl1% [247,2] Sodium Chlorideqs Purified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye. How many g of NaCl needed? Using Freezing point method.

202 247 gL -1 = o C*1.8 10gL -1 - x o C o C o C ( o C)= o C ( o C)= o C 0.52-

203 0.270 g NaCl Reference 0.27 g freezes at-0.52 o C x g NaCl o C g NaCl

204 P22 Rx Oxytetracycline HCl 0.05 [ Oxytetracycline HCl 0.05 [ 497,2] Chlorobutanol0.1 [177,1] NaCl qs Purified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye. How many mg of NaCl needed?

205 “E” for Oxytetracycline HCl 497gL -1 * gL -1 *1.8 = g Oxy* = g NaCl

206 “E” for Chlorobutanol 177gL -1 * gL -1 *1.0 = g Chl* = g NaCl

207 Oxytetracycline HCl g Chlorobutanol g g g NaCl Reference g NaCl g

208 P23 Rx Tetracaine HCl 0. 5% [ Tetracaine HCl 0. 5% [ 301,2] [iso]Sol. Epineph. Bitart.10.0 [ [iso]Sol. Epineph. Bitart.10.0 [ 333,2] Boric acid qs Purified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye. How many g of Boric acid needed?

209 “Eb” for Tetracaine HCl 301gL -1 * gL -1 *1.8 = g Tet.* = g B.A g*20 mL/1000mL g B.A. for 20 mL Reference

210 0.2946g Boric acid g B.A.

211 P24 Anhyd.NaH 2 PO g[120,2] Anhyd.Na 2 HPO g[142,3] NaClqs Purified Water ad1000 mL Label: Isotonic buffer sol.,pH6.5 How many g of NaCl needed?

212 “E” for Anhyd.NaH 2 PO gL -1 * gL -1 *1.8 = g Mono* 0.49= g NaCl

213 “E” for Anhyd.Na 2 HPO 4 = g Di* 0.595= 1.69 g NaCl 142 gL -1 * gL -1 *2.6

214 Anhyd.NaH 2 PO g NaCl Anhyd.Na 2 HPO g NaCl g NaCl 9.0 g NaCl Reference g NaCl g NaCl

215 P25 How many g of anhydrous Dextrose needed in preparing 1 L of a 0.5% isotonic Ephedrine Sulfate [429,3] Nasal spray?

216 “Ed” for Ephedrine Sulfate 429 gL -1 * gL -1 *2.6 = g Ephd.* 1.09= 5.45 g Dex. 50 gL -1 Anhydrous Dex.Ref gL -1 Dex g Dextrose

217 P26 Ephedrine Sulfate 1% [429,3] Chlorobutanol 0.5%[177,1] Purified Water ad Make isoton. sol. & Buffer at 6.5 Sig. Nose drop. How many mL of a buffer & mL of water should be used?

218 “E” for Ephedrine Sulfate 429gL -1 * gL -1 *2.6 = g Eph* = g NaCl

219 V-Value for Ephedrine Sulfate g NaCl 100 mL H 2 O 0.9 g NaCl x mL mL of water will make 1 g of Ephedrine Sulfate isotonic.

220 “E” for Chlorobutanol 177gL -1 * gL -1 *1.0 = g Ch* = g NaCl

221 V-Value For Chlorobutanol g NaCl 100 mL H 2 O 0.9 g NaCl x mL x mL= mL of water will make 0.5 g of chlorobutanol isotonic.

222 Ephedrine Sulf mL of water Chlorobutanol mL of water mL of water mL of isotonic buffer solution.

223 P27 Oxytetracycline HCl 0.5% [ Oxytetracycline HCl 0.5% [ 497,2] [iso] Tetracaine HCl Sol. 2% 15 ml NaCl qs Purified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye. How many mL of NSS needed?

224 “E” for Oxytetracycline HCl 497gL -1 * gL -1 * = 01177g NaCl g Oxy* =

225 0.135 g NaCl Reference in only 15 mL g NaCl Equivalent g. NaCl 0.9 g NaCl100 mLx mL g. NaCl 13.0 mL of NSS

226 Determine if the following commercial products are Hypotonic, isotonic, or Hypertonic: a- An ophthalmic sol. 40 mgmL -1 of Cromolyn Sodium [512,2]& 0.01% of Benzalkomiun chloride [ a- An ophthalmic sol. 40 mgmL -1 of Cromolyn Sodium [512,2]& 0.01% of Benzalkomiun chloride [ 360,2] in purified water.

227 “E” for CromolynSodium 512gL -1 = o C*1.8 40gL -1 = -X o C o C >-0.52 <-0.52 HYPO ISO-T. HYPER

228 A parenteral infusion containing 20% (w/v) of mannitol. 182gL -1 = o C* gL -1 = -X o C >-0.52 <-0.52 HYPO ISO-T. HYPER FP ”D” Mannitol

229 A 500-mL large volume parenteral containing D5W (5% w/v of anhydrous dextrose in sterile water for injection). 180gL -1 = o C*1.0 50gL -1 = -X o C >-0.52 <-0.52 HYPO ISO-T. HYPER

230 A FLEET saline enema containing 19 g of monobasic sodium phosphate (monohydrate) and 7 g of dibasic sodium phosphate (heptahydrate) in 118 mL of aqueous solution.

231 Monobasic sodium phosphate (monohydrate) (138, 2) 138gL -1 = o C* gL -1 = -X o C >-0.52 <-0.52 HYPO ISO-T. HYPER -3.9

232 Dibasic sodium phosphate (heptahydrate) (268,3) 268gL -1 = o C* gL -1 = -X o C >-0.52 <-0.52 HYPO ISO-T. HYPER -1.07

233 For agents having the following sodium chloride equivalents, calculate the percentage concentration of an isotonic solution: (A) % 0.20 = 4.5% (b) % 0.32 = 2.81% (c) % 0.61 = 1.48%

234 How many mL each of purified water and an isotonic sodium chloride solution should be used to prepare 30 mL of a 1% w/v isotonic solution of fentanyl citrate (E = 0.11)? P30 1 g * 3 mL/100 mL = 0.3 g fentanyl citrate 0.3 g fentanyl citrate* 0.11= g NaCl g NaCl* 100 mL/0.9 g NaCl= 3.66 mL H 2 O 30 mL mL H 2 O= mL N.S.

235 Calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following: (a) Antipyrine [ 188, 1] 58.5*1/(188*1.8) = g * 0.3 = g NaCl g NaCl* 100 mL / 0.9 g NaCl = 5.76 mL H 2 O

236 Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following: (b) Chlorobutanol [177, 1] 58.5*1/(177*1.8) = g * 0.3 = g NaCl g NaCl* 100 mL / 0.9 g NaCl = 6.16 mL H 2 O

237 Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following: (c) ephedrine sulfate [429, 3] 58.5*2.6/(429*1.8) = g * 0.3 = g NaCl g NaCl* 100 mL / 0.9 g NaCl = 6.56 mL H 2 O

238 Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following: (d) silver nitrate [170, 2] 58.5*1.8/(170*1.8) = g * 0.3 = g NaCl g NaCl* 100 mL / 0.9 g NaCl = mL H 2 O

239 Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following: (e) zinc sulfate [288, 2] 58.5*1.4/(288*1.8) = g * 0.3 = g NaCl g NaCl* 100 mL / 0.9 g NaCl = mL H 2 O


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