A Simulated Body Fluid (SBF) is an artificially prepared solution that is used b

Question

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 Calculate the Molarity of the various salts listed above, and the number of milli-equivalents per liter of the various ions present (Na^+, K^+, Mg^++, Cl^-, HCO_3^-, HPO_4^-2) [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-]

Explanation / Answer

Step 1: To find no. of moles of all the components

Molar mass of NaCl,M1 = 58.44g/mol Molar mass of sodium bicarbonate,M2 = 84.007g/mol Molar mass of KCl,M3 = 74.55g/mol Molar mass of Potassium phosphate dibasic trihydrate,M4 = 228.23g/mol Molar mass of Magnesium chloride hexahydrate, M5 = 203.30g/mol

Mass of NaCl,m1 = 8.035g Mass of sodium bicarbonate,m2 = 0.355g Mass of KCl,m3 = 0.255g Mass of Potassium phosphate dibasic trihydrate,m4 = 0.231g Mass of Magnesium chloride hexahydrate, m5 = 0.311g

No. of moles,n = mass of compound/ molar mass of compound

Moles of NaCl,n1 = 8.035g / (58.44g/mol) = 0.1375 mol Moles of sodium bicarbonate,n2 = 0.355g /(84.077g/mol) = 0.0042 mol Moles of KCl,n3 = 0.255g/(74.55g/mol) = 0.0034 mol Moles of Potassium phosphate dibasic trihydrate,n4 = 0.231g/(228.23g/mol)= 0.0010 mol Moles of Magnesium chloride hexahydrate, n5 = 0.311g /(203.30g/mol) = 0.0015 mol

Molarity = no. of moles / volume of solution in litres

Molarity of NaCl,c1 = 0.1375 mol / 1L = 0.1375M Molarity of sodium bicarbonate,c2 = 0.0042 mol / 1L = 0.0042M Molarity of KCl,c3 = 0.0034 mol/ 1L = 0.0034M Molarity of Potassium phosphate dibasic trihydrate,c4 = 0.0010 mol/1L = 0.0010M Molarity of Magnesium chloride hexahydrate, c5 = 0.0015 mol/ 1L = 0.0015M      

To find molarity of HCl , we will use C1V1 = C2V2   Conc of stock HCl solution ,C1 = 1M Volume of stock HCl solution taken, V1 = 35mL Volume of the solution, V2 = 1L = 1000mL C2 = C1V1 / V2 = 1M X(35mL) / 1000mL = 0.035M   

Molarity of HCl = 0.035M

equivalent weight = molecular weight /valence  

Valence of NaCl, v1 = 1 Valence of NaHCO3, v2 = 1 Valence of KCl,v3 = 1 Valence of potassium phosphate dibasic trihydrate, v4 = 2 Valence of MgCl2.6H2O , v5 = 2 Valence of HCl , v6 = 1

equivalent weight of NaCl,w1 = 58.44g equivalent weight of sodium bicarbonate,w2 = 84.007g equivalent weight of KCl,w3 = 74.55g equivalent weight of Potassium phosphate dibasic trihydrate,w4 = 228.23/2 = 114.115g equivalent weight of Magnesium chloride hexahydrate, w5 = 203.30/2 = 101.65g

No. of equivalents = mass of Substance / equivalent weight of substance

No. of equivalents of NaCl = 8.035/58.44g = 0.1375eq No. of equivalents of sodium bicarbonate= 0.355g/84.007g = 0.0042eq No. of equivalents of KCl = 0.255g/74.55g = 0.0034eq No. of equivalents of Potassium phosphate dibasic trihydrate =0.231g/114.115g = 0.002eq No. of equivalents of Magnesium chloride hexahydrate = 0.311g/ 101.65g = 0.003eq

1 eq = 1000meq

meq/L of NaCl = (0.1375eq *1000)/L = 137.5meq/L meq/L of sodium bicarbonate= 4.2meq/L meq/L of KCl = 3.4meq/L meq/L of Potassium phosphate dibasic trihydrate = 2meq/L meq/L of Magnesium chloride hexahydrate = 3meq/L

For HCl, 1 mol = 1 equivalent , so 0.035 moles/L = 0.035eq/L of HCl = 0.035eq * 1000meq /L=35meq/L 1eq meq/L of HCl = 35meq/L

                 

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