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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
Chapter 15, Problem 15-A(choose chapter or problem)
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. 15-B
Questions & Answers
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. 15-B
ANSWER:Step 1 of 4
Given that:
20 mL solution
And, 0.005 00 M in 1 M HCl was titrated with 0.020 0 M to give and.
Hence 1 mol of requires 2 moles of .