I am unsure of what number from #3 I am using for #4 to find [FeSCN2+]eq. 4. In
ID: 519844 • Letter: I
Question
I am unsure of what number from #3 I am using for #4 to find [FeSCN2+]eq. 4. In Part C, you prepare four solutions (starting concentrations are given below) in test tubes and measure the Absorbance after the solutions have reached equilibrium. Shown below is an example of just one solution (Absorbance data also below). Using this data and the standard curve from number 3. (on previous page), calculate the Kf Starting Concentrations 0.00200 M Fe (NO)3 0.0020 M KSCN (mL) H20 (mL) Test tube (malo 5.0 2.0 3.0 Absorbance Data Test tube Absorbance at FeSCN at Equil Kr 0.118Explanation / Answer
Use the linear equation to find the equilibrium concentration of [Fe(SCN)2+]. Put y = 0.118 and obtain x.
0.118 = 2417.5x + 0.0483
===> 2417.5x = 0.0697
===> x = 2.883*10-3
The equilibrium concentration of Fe(SCN)2+, i.e., [Fe(SCN)2+] = 2.883*10-5 M (ans).
Total volume of solution = 10.0 mL.
Moles of Fe(SCN)2+ at equilibrium = (10.0 mL)*(2.883*10-5 mol/L) = 2.883*10-4 mmole.
Moles of Fe(NO3)3 added initially = (3.0 mL)*(0.00200 mol/L) = 6.0*10-3 mmole.
Moles of Fe(NO3)3 retained at equilibrium = [moles of Fe(NO3)3 added] – [moles of Fe(SCN)2+] formed at equilibrium = (6.0*10-3 – 2.883*10-4) mmole = 5.7117*10-3 mmole.
Molar concentration of Fe(NO3)3 retained at equilibrium = (5.7117*10-3 mmole)/(10.0 mL) = 5.7117*10-4 M.
Moles of KSCN added initially = (2.0 mL)*(0.00200 mol/L) = 4.0*10-3 mmole.
Moles of KSCN retained at equilibrium = [moles of KSCN added] – [moles of Fe(SCN)2+] formed at equilibrium = (4.0*10-3 – 2.883*10-4) mmole = 3.7117*10-3 mmole.
Molar concentration of KSCN retained at equilibrium = (3.7117*10-3 mmole)/(10.0 mL) = 3.7117*10-4 M.
The equilibrium reaction is
Fe3+ (aq) + SCN- (aq) ------> Fe(SCN)2+
The equilibrium constant is given as
Kf = [Fe(SCN)2+]/[Fe3+][SCN-] = (2.883*10-5 M)/(5.7117*10-3 M).(3.7117*10-3 M) = 1.619 M-1 1.62 M-1 (ans).
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