For one of the solutions you make up for this week\'s experiment, 7.00 mL of 2.0

Question

For one of the solutions you make up for this week's experiment, 7.00 mL of 2.0E-3 M iron(III) nitrate solution is mixed with 2.00 mL of 2.0E-3 M potassium thiocyanate solution, 3.00 mL of 0.500 M nitric acid solution, and 8.00 mL of distilled water for a total volume of 20.00 mL. See line 3 in Table 1. What is the initial concentration of Fe3+ ions in this solution, in other words, what is the concentration of Fe3+ ions before any reaction with the thiocyanate? Initial concentration of Fe3+ = (answer)M

Calculate the mole fractions of Fe3+ and SCN- in this solution with respect to the total number of moles of Fe3+ and SCN-. This means, of course, that these two mole fractions should add up to 1.00. Mole fraction of Fe3+ =(answer)

Mole fraction of SCN- =(answer)

Suppose that the mole fraction of Fe3+ in the iron(II) - thiocyanate solution was plotted against the absorbance of the solution at 450 nm. Suppose also that the maximum absorbance was established at a mole fraction of Fe3+ equal to 0.500 by plotting the two best straight lines through the data points. What is the stoichiometry of the complex formed between the Fe3+ and the thiocyanate? Enter the answer as a whole number ratio; whole numbers separated by a colon (for example, 2:3, indicating a stoichiometry Fe2(SCN)3). Stoichiometry of the complex formed between Fe3+ and SCN- =(answer)

Explanation / Answer

0.007L x 0.002M Fe(NO3)2 = 1.4x10^-5 moles Fe3+ in Fe(NO3)3
moles Fe3+ / total volume of the solution = molarity of Fe 3+
1.4x10^-5moles Fe3+ / 0.02L = 7x10^-4M Fe3+ present

mole fraction Fe3+ = moles Fe3+ / total moles present
moles SCN- = 0.002L x 2x10^-3M = 0.4x10^-5moles SCN-
1.4x10^-5moles Fe3+ / (1.4x10^-5 + 0.4x10^-5) = 0.778
mole fraction SCN- = 0.222

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