Please provide the steps for the solution. I am particularly confused with what

Question

Please provide the steps for the solution. I am particularly confused with what I'm supposed to subtract the total moles of HCl with. I think the concentration of the solution can be determined via M1V1=M2V2 but not sure

back titration method was used to titrate a sample of a polyelectrolyte (a weak base). A 2.056 g sample of the polyelectrolyte is titrated with 100.0 ml of 0.1021 M HCl by dissolving the polyelectrolyte in the strong acid solution. The solution was then filtered and 25.00 mL of the filtrate was removed and titrated with NaOH. It required 30.45 ml of 0.04857 M NaOH to reach the equivalence point. What is the formula weight of the polyelectrolyte in g/mol.

Explanation / Answer

Weight of polyelectrolyte/weak base = 2.056 g

Initial Volume of HCl, V = 100 ml = 0.1 lt

Initial Molarity of HCl, M = 0.1021 M

Initial no. Of moles of HCl = molarity of HCl x vol of HCl in lts = M x V = 0.1021 moles/lt x 0.1 lt = 0.01021 moles

Molarity of NaOH, M2 = 0.04857 M

Volume of NaOH, V2 = 30.45 ml = 0.03045 lts

No. Of moles of NaOH = molarity x vol of NaOH in lts = M2V2

= 0.04857 mol/lts x 0.03045 lts

= 0.00148 moles of NaOH

NaOH + HCl -----> NaCl + H2OO

As molar ratio of NaOH and HCl is 1:1

So the number of moles of HCl back titrated with NaOH = No. Of moles of NaOH = 0.00148 moles

No. Of moles of HCl in 20 ml = 0.00148 moles

No. Of moles of HCl in 100 ml = 0.00148 moles x100 ml/20 ml = 0.0074 moles

No. Of HCl consumed = initial no. Of moles of HCl -no. Of HCl moles in 100 ml = 0.01021 moles - 0.0074 moles = 0.00281 moles

Let us consider polyelectrolyte is a monoacidic weak base = MOH

MOH + HCl ---------> MCl + H2O

As molar ratio of polyelectrloyte and HCl is 1:1

So, No. Of moles of Polyelectrolyte /weak base = No. Of moles of HCl consumed = 0.00281 moles = weight of polyelectrolyte/molecular weight

2.056 gms/molecular weight = 0.00281 moles

Molecular weight = 2.056 gms/0.00281 moles = 731.673 g/mol

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