2. A Simulated Body Fluid (SBF) is an artificially prepared solution that is use

Question

2. A Simulated Body Fluid (SBF) is an artificially prepared solution that is used by pharmacists to studying in-vitro the dissolution kinetics of drugs in body fluids such as plasma (blood), interstitial fluid (ISF), Cerebrospinal Fluid (CSF) etc. Various formulations are available in literature. One particular formulation is listed below Reagents for Lof SBF 8.035 g Sodium Chloride 0.355 g Sodium Bicarbonate 0.225 g Potassium Chloride potassium phosphate dibasic trihydrate 0.231 0.311 magnesium chloride hexahydrate 1 M Hydrochloric Acid 35 ml 965 ml DI Water Calculate the (a) Molarity of the various salts listed above, and (b) the number of milli-equivalents per liter of the various ions present (Nat, K, Mg Cl, HCO, HPO4 [make sure you account for the presence of HCl in the mixture. You can assume that all the HCl is completely dissociated into H+ and Cl-J

Explanation / Answer

(a) Sodium chloride(NaCl): Mass of NaCl = 8.035 g

Molecular mass of NaCl = 58.44 g/mol

Hence moles of NaCl = mass / molecular mass = 8.035 g / 58.44 g/mol = 0.1375 mol

Volume of SBF = 1 L

Hence molarity of NaCl = moles of NaCl / Volume(L) = 0.1375 mol / 1L = 0.1375 M (answer)

Sodium bicarbonate(NaHCO3): Mass of NaHCO3 = 0.355 g

Molecular mass of NaHCO3 = 84.0 g/mol

Hence moles of NaHCO3 = mass / molecular mass = 0.355 g / 84.0 g/mol = 0.00423 mol

Volume of SBF = 1 L

Hence molarity of NaHCO3 = moles of NaHCO3 / Volume(L) = 0.00423 mol / 1L = 0.00423 M (answer)

Potassium chloride(KCl): Mass of KCl = 0.225 g

Molecular mass of KCl = 74.55 g/mol

Hence moles of KCl = mass / molecular mass = 0.225 g / 74.55 g/mol = 0.00302 mol

Volume of SBF = 1 L

Hence molarity of KCl = moles of KCl / Volume(L) = 0.00302 mol / 1L = 0.00302 M (answer)

Potassium phosphate dibasic trihydrate(K2HPO4.3H2O): Mass of K2HPO4.3H2O = 0.231 g

Molecular mass of K2HPO4.3H2O = 228.22 g/mol

Hence moles of K2HPO4.3H2O = mass / molecular mass = 0.231 g / 228.22 g/mol = 0.00101 mol

Volume of SBF = 1 L

Hence molarity of K2HPO4.3H2O = moles of K2HPO4.3H2O / Volume(L) = 0.00101 mol / 1L = 0.00101 M (answer)

Magnesium chloride hexahydrate(MgCl2.6H2O): Mass of MgCl2.6H2O = 0.311 g

Molecular mass of MgCl2.6H2O = 203.30 g/mol

Hence moles of MgCl2.6H2O = mass / molecular mass = 0.311 g / 203.30 g/mol = 0.00153 mol

Volume of SBF = 1 L

Hence molarity of MgCl2.6H2O = moles of MgCl2.6H2O / Volume(L) = 0.00153 mol / 1L = 0.00153 M (answer)

(b) For NaCl, equivalent mass = molecular mass

Hence number of equilivants of Na+ = number of moles of NaCl = 0.1375 eqv

Hence milliequivalents of NaCl =0.1375 eqv x ( 1 millieqv / 10-3 eqv) = 137.5 milliequivalents

Hance millieqv of Cl- = 137.5 milliequivalents

millieqv of Na+ due to NaCl = 137.5 milliequivalents

For NaHCO3, equivalent mass = molecular mass

Hence number of equilivants of NaHCO3 = number of moles of NaHCO3 = 0.00423 eqv

Hence milliequivalents of NaHCO3 =0.00423 eqv x ( 1 millieqv / 10-3 eqv) = 4.23 milliequivalents

Hence milliequivalents of HCO3- = 4.23 milliequivalents (answer)

Similarly we can calculate milliequivalents for the rest as

milliequivalents of Na+ = millieqv in NaCl + millieqv in NaHCO3

= 137.5 meqv + 4.23 meqv = 141.73 meqv (answer)

milliequivalents of K+ = millieqv in KCl + millieqv in K2HPO4.3H2O

= 3.02 meqv + 2x1.01 meqv = 5.04 meqv (answer)

milliequivalents of Mg2+ = millieqv in MgCl2.3H2O = 1.53 meqv (answer)

millieqv of Cl- = millieqv in NaCl + millieqv in KCl + millieqv in MgCl2.3H2O + millieqv in 35mL of1M HCl

= 137.5 meqv + 3.02 meqv + 2x1.53 meqv + 35 meqv

= 178.5 meqv (answer)

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.