A 20.0-mL solution of 0.005 00 M Sn 2+ in 1 M HCl was titrated with 0.020
Question:
A 20.0-mL solution of 0.005 00 M Sn2+ in 1 M HCl was titrated with 0.020 0 M Ce4+ to give Sn4+ and Ce3+. Calculate the potential (versus S.C.E.) at the following volumes of Ce4+: 0.100, 1.00, 5.00, 9.50, 10.00, 10.10, and 12.00 mL. Sketch the titration curve.
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To calculate the potential of the solution at different volumes of Ce4 we need to know the concentrations of all species present in the solution at ea...View the full answer
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