Step 1 Finding the Initial Number of Moles of Each Reactant. This requires relat

Question

Step 1 Finding the Initial Number of Moles of Each Reactant. This requires relating the vol- umes and concentrations of the reagent solutions that were mixed to the numbers of moles of each reactant species in those solutions. By the definition of the molarity, M A, of a species no, moles A M,_noT tersorsolution,T no moles A-M.xv or Using Equation 3, it is easy to find the initial number of moles of Fe and SCN. For each solution the volume used was 10.0 mL., or 0.0100 L. The molarity of each of the solutions was 2.00 × 10-1 M, so Mre" = 2.00 × 10.) M and Mo = 2.00 × 10-3 M. Therefore, in the reagent solutions, we find that initial no. moles Fe.= Mr.x V= 2.00 x 10-1 M x 0.01 00 L-20.0 x 10 moles initial no. moles SCN-= Mscx × V-2.00 x 10-, M x 0 0 100 L-20.0 x 10 moles Step 2 Finding the Number of Moles of Product Formed. Here again we can use Equation 3 to advantage. The concentration of FeSCN was found to be 1.50x 10 M at equilibrium. The volume of the mixture at equilibrium is the sam of the two volumes that were mixed, and is 20.0 mL, or 0.0200 L. So no, mo les FeSCN. . MhMN. × V. 1.50 x 10-4 M x 0.0200 L . 3.00 x 10-6 moles The number of moles of Fe and SCN that were ased up in producing the FeSCN must also both be equal to 3.00 x 10* moles since, by Equation 1, it takes one mole Fe and one mole SCN to make each mole of FeSCN Step 3 Finding the Number of Moles of Each Reactant Present at Equilibrium. In Step I we determined that initially we had 20.0 x 10'+ moles Fe, and 20.0 × 10-6 moles SCN' pres- ent. In Step 2 we found that in the reaction 3.00 x 10-6 moles Fe" and 3.00 × 10-6 moles SCN were used up. The number of moles present at equilibrium must equal the number we started with minus the number that reacted. Therefore, at equilibrium no. moles at equilibrium = initial no. moles-no, moles used up equil, no, moles Fe''. 20.0 × 10-6-3.00 x 10-6-17.0 x 10-6 moles equil. no. moles SCN-= 20.0 x 10-_ 3.00 x 10'.-17.0 x10-moles Find the Concentrations of All Species at Equilibrium. Experimentally, we obtained the concentration of FeSCN°, directly. [FeSCN2+ 1.50 x 10-4 M. The concentrations of Fe" and SCN follow from Equation 3. The number of moles of each of these species at equi- librium was obtained in Step 3. The volume of the mixture being studied was 20.0 mL., or 0.0200 L. So, at equilibrium Step 4 no. moles Fe_ = 17.0x10"moles. 8.50 x 10"M 0.0200 L [Fe'*)=M,..= Fe"]-Mvolume of solution no, moles Fe'. no. moles SCN 17.0x10 moles = 8.50×10-4 M [SCN ]. MscN -volume orsolution = 0.0200 L Step 5 Finding the Value of K, for the Reaction. Once the equilibrium concentrations of all the reactants and products are known, one needs merely to substitute into Equation 2 to deter- mine FeSCN ] 1.50×10-4 - 208

Explanation / Answer

Method 1

use the graph of absorbance vs concentratiion of FeSCN2+ (standard solutions) to find the value of [FeSCN2+] for unknown test solutions.

For mixtures 1,2 let me locate the value from the graph provided.

mix 1

absorbance=0.09    [FeSCN2+]=0.19*10^-5 M   moles of FeSCN2+=0.19*10^-5 M *(total volume of solution)=0.19*10^-5 M *(5ml+1ml+0.004 ml)=0.19*10^-5 M *6.004 ml=0.19*10^-5 M *6.004*10^-3 L=1.14*10^-8 moles

mix 2 abs=0.24    [FeSCN2+]=0.5*10^-5 M =0.5*10^-5 M *7.003*10^-3 L=3.5*10^-8 moles

Initial no of moles

mix 1

initial moles of Fe(NO3)3 =M*V=2.0*10^-3 mol/L * 0.005 L=1*10^-5 moles

initial moles of KSCN=M*V=2.0*10^-3 mol/L * 0.001 L=2*10^-6 moles

equilibrium moles of Fe(NO3)3= initial moles -moles of FeSCN2+ =1*10^-5 moles -1.1*10^-8 mol=1*10^-5 -0.0011*10^-5=0.999*10^-5 moles

equilibrium conc=eqm moles/total volume =0.999*10^-5 moles/6.004 *10^-3L=1.6*10^-3 M

Similarly

eqm moles of KCNS=2*10^-6 moles-1.1*10^-8 mol=2*10^-6 moles

eqm conc =2.0*10^-6 moles/6.004*10^-3 L=3.3*10^-4 M

mix 2

initial moles of Fe(NO3)3 =M*V=2.0*10^-3 mol/L * 0.005 L=1*10^-5 moles

initial moles of KSCN=M*V=2.0*10^-3 mol/L * 0.002 L=6*10^-6 mol

equilibrium moles of Fe(NO3)3= initial moles -moles of FeSCN2+ =1*10^-5 moles -3.5*10^-8 moles mol=1*10^-5 -0.0035*10^-5=1*10^-5 moles

equilibrium conc=eqm moles/total volume =1*10^-5 moles/7.003 *10^-3L=1.4*10^-3 M

eqm moles of KSCN=6*10^-6 mol-3.5*10^-8 moles mol=6*10^-6 moles

eqm conc of KSCN=6*10^-6///7.003 *10^-3L=8.6*10^-3 M

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.