Question

Part A As a technician in a large pharmaceutical research firm, you need to produce 350. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.92. The pKa of H2PO4 is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.) Express your answer to three significant digits with the appropriate units. ► View Available Hint(s) .: HÅR O 2 ? Volume of KH2PO4 needed = Value Units Submit

Answer 1

The Ka values for the phosphoric acid are The phosphoric acid has the following ionizations H3PO4 = H2PO4 + H+ H2PO4 = HPO4-2 + H+ pka2 = 7.21 HPO4-27 POA-3 + H+ Since the given pH (6.92) is nearer to second ionization. The phosphate buffer must be prepared by H2PO4-1 and HPO4-2 [base] pH = pk +log [acid] =6.92 = 7:21+ log in [H,PO, log . [HPO,"] _-0.29 10% [H,PO, ] [HPO ) - 10-0.29 = 0.512861 [H,PO] [HPO, 2]=0.512861x[H,PO, ] The conc. of buffer =1 M That is [HPO, ?]+[H,PO,-]=1 0.512861x[H,PO, ]+[H,PO - ]=1 1.512861x[H,PO,"]=1 [H,PO,+1= 1512861 = 0.661 [HPO, ?]=0.512861x[H,PO, ]=0.512861x0.661=0.339 [H,PO, ]=0.661M [HPO, ?]=0.339M Volume of the buffer solution = 350 mL = 0.350L Number of moles of H,PO, present in the solution = M*V = 0.661M 0.350 L = 0.23135 mol Number of moles of HPO, ² present in the solution = M*V = 0.339Mx0.350 L = 0.11865 mol moles Volume of KH,PO, solution == Molarity moles Volume of K2HPO4 solution = Molarity |Volume of KH,PO, solution = 231mL 0.23135 mol P=0.23135 L = 231 mL IM 0.11865 mol = 0.11865 L = 119 mL IM

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The bicarbonate buffer system is CO,+H,02H,CO, 2HCO +H* pH of the buffer is determined by the Henderson equation, CO, +H20 FH,CO3 + HCO3 +H* pH = pKa +log (HCO3] 10% [H,CO2] According to Henry law, [H,CO2] = Ky * Pco =0.03 (mmol/L)/mmHg x Pco, So, the modified Henderson equation is [HCO, pH = pKa+log - 18 0.03 (mmol/L)/mmHg x Pco, Given that [HCO3-]= 24 mM = 24 mmol/L Pcos = 31.0 mm Hg pKa = 6.1 [HCO3-) pH = pka + log 0.03 (mmol/L)/mmHg x P coz 24 mmol/L pH = 6.1+ log- 0.03 (mmol/L)/mmHg x 31.0 mm Hg pH =6.1+1.411728 = 7.511728 pH of the blood = 7.51