During coulometric titration of an organic compound,XH 2 , with an iodide soluti

Question

During coulometric titration of an organic compound,XH2 , with an iodide solution, the time to reach the endpoint was 30.0s at 5.0 mA. ( The XH2 molecular weightwas 85.00 g/mol). F = 96490 C/mol The reactions to consider are: 2I- = I2 + 2e- I2 + XH2 = X + 2HI (a) What was the contents of XH2(mg) in the sample? (b) Assuming that the volume of the solution was 500 mL, whatwas the molar concentration of XH2? During coulometric titration of an organic compound,XH2 , with an iodide solution, the time to reach the endpoint was 30.0s at 5.0 mA. ( The XH2 molecular weightwas 85.00 g/mol). F = 96490 C/mol The reactions to consider are: 2I- = I2 + 2e- I2 + XH2 = X + 2HI (a) What was the contents of XH2(mg) in the sample? (b) Assuming that the volume of the solution was 500 mL, whatwas the molar concentration of XH2? I2 + XH2 = X + 2HI (a) What was the contents of XH2(mg) in the sample? (b) Assuming that the volume of the solution was 500 mL, whatwas the molar concentration of XH2?

Explanation / Answer

           a)     The reactions to consider are:                 2I- = I2 + 2e-                 I2 + XH2 = X + 2HI        a)     The reactions to consider are:                 2I- = I2 + 2e-                 I2 + XH2 = X + 2HI                 I2 + XH2 = X + 2HI               Molecular weight of organic compound ,XH2    M.W = 85g/mol            current passed through the sampleXH2            i = 5.0mA                                                                                       = 5.0*10-3 A              time taken for passingcurrent                         t = 30.0 s              And faraday constant F = 96490 C/mol                                                     =96490 A.s /mol           But we know the relation between mass of sample and current passedon sample is                     m = Z*i*t , where Z = molar mass / ne*F        Now substitute all of theabove values in this equation we get     mass of organic sampleXH2    m = [(85g/mol ) / (2*96490A.s/mol)]*(5.0*10-3 A)*(30.0s)                                                        = 6.61*10-5 g b) Here volume of the solution V = 500.0mL                                                     = (500/1000)L or 0.5 L              and we know molecular weight of sample M.W = 85.0g/mol                                 mass ofsample                       m =  6.61*10-5 g               number of moles of sample n = mass / molecular weight                                                            =( 6.61*10-5 g )/ (85.0 g/mol)                                                            = 7.78*10-7 mol               Molar concentration of XH2    M = moles /Volume in L                                                                     = 7.78*10-7 mol / 0.5L                                                                     = 1.56*10-6 M                                                           

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