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A protein 13.2 Weight-percent N. A .500 ml Aliquot of protein solution was digested and liberated...

A protein 13.2 Weight-percent N. A .500 ml Aliquot of protein solution was digested and liberated NH3 was distilled into a beaker containing 10.0mL of .01380 M HCl. Unreacted HCl needed 1.54 mL of .0108 M Ba(OH)2 for titration. What was the concentration of protein (mg/mL)

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Expert Solution

HCl moles = M x V = 0.0138 x 10/1000 = 0.000138

Ba(OH)2 moles = Mx V = 0.0108 x 1.54/1000 = 0.000016632

1 Ba(OH)2 + 2Hcl --> BaCl2 + 2H2O

hence Hcl moles reacted with Ba(OH)2 = 2 x Ba(OH)2 moles = 2 x 0.000016632 = 0.000033264

Unreacted HClmoles = 0.000033264

HCl moles reacted = intial Hcl moles - unreacted Hcl moles

                        = 0.000138-0.0000033264

                              = 0.0001047364

HCl and NH3 react in 1:1 ratio hence NH3 moles = 0.0001047364

N moles = 0.0001047364   ( since 1 NH3 has 1 N)

N mass = moles x molar mass of N = 0.0001047364 x 14 = 0.0014663 g

N mass % = 13.2 which means we have 13.2 g N per 100 g protein

hence for 0.0014663 g N we have protein mass = ( 0.0014663/13.2) x 100 = 0.0111 g = 11.1 mg

solution = 500 ml

protein concentration in mg / ml = (11.1/500) = 0.0222

queen_honey_blossom answered 1 month ago
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