7. If 1.0 mL of 0.002 M Fe3 (aq), 1.0 mL 0.002 M scN (aq) and 8.0 mL of H20 are

Question

7. If 1.0 mL of 0.002 M Fe3 (aq), 1.0 mL 0.002 M scN (aq) and 8.0 mL of H20 are mixed together as the initial solution, what will be the initial concentration of SCN (aq)? 8. According to the equation Fe3+(aq) SCN (aq) [Fe(SCN)12 (aq), If 1.0 mL of 0.002 M Fes (aq), 1.0 mL 0.002 M SCN (aq) and 8.0 mL of H2O are mixed together as the initial solution, what will be the initial concentration of [Fe(SCN)21 (aq)? the calibration part of the experiment (Part 1), the reactants are combined with one as a limiting reagent. According to the equation Fe3 (aq) SCN (ag) e [Fe(SCN)? (aq), if 1.0 mL of 0.2 M Fes (aq), 1.0 mL 0.002 M SCN and 69.0 mL of H2O are mixed together as the initial solution, what will be the equilibrium concentration of Fe3 (aq) 10. In the calibration part of the experiment (Part 1), the reactants are combined with one as a limiting reagent. According to the equation Fes SCN (aa), if the inital 69.0 mL of H20 are mixed together as solution, what will be the equilibrium concentration of SCN (aq)?

Explanation / Answer

7.

First calculate the moles of SCN- as follows:

Number of moles = volume * molarity

= 1/1000 * 0.002

= 2*10^-6 mole SCN-

Total volume = 1.0 ml+1.0 ml+8.0 ml

= 10 ml

= 0.010 L

New molarity of SCN-= number of moles / volume in L

= 2*10^-6 mole SCN- / 0.010 L

=2.0*10^-4 M

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