TABLE of TUBE DILUTED Fe 1 M KSCN 5.00 5.00 0.1 MHNO 1.00 2.00 3.00 4.00 3.00 2.
ID: 704494 • Letter: T
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TABLE of TUBE DILUTED Fe 1 M KSCN 5.00 5.00 0.1 MHNO 1.00 2.00 3.00 4.00 3.00 2.00 5.00 13 Determination of Ke for an Equilibrium System Pre-laboratory Assignment Name Section 1. Calculate [Fe'Jaist for test tubes 1-3. Show your work here for one of the tubes 2. A student performed a lab similar to this one in an attempt to find Ke for the equilibrium system X')+Y Z. In one trial, the student combined 5.00 mL of 0.10 M X with 10.00 mL of 0.20 M Y' in a test tube. When the resulting solution was analyzed with a spectrophotometer set at 600 nm, 4.00% of the light passed through the solution. For this equilibrium system, the Beer's law constant, ß, is known to be 924 L mol Calculate the initial molarity of each reactant added to the test tube at the moment of mixing, [Xnir and [YIni.(HINT: these are dilution calculations.) A. B. Calculate the value of A, then find [Zloqua. C. Set up an LCE, table to find [Xhqiand [Y)equil. D. Find the value of Ke for this equilibrium system. revised 4/6/2016 DN.Explanation / Answer
2)
ans)
from the above information given that
A)
concentration of X+ =0.1M, volume= 5ml
moles of X+ added= concentraion in M* Volume in L= 0.1*5/1000 =0.0005
similarly, moles of Y- = 0.2*10/1000 = 0.002
when combined, volume= 10+5= 15ml= 15/1000L= 0.015L
after mixing concentrations : X+= 0.0005/0.015
X=0.0333M and
Y=0.002/0.015
Y=0.1333M
B)
since 4%of light passes through the solution , A= -log (4/100)= 1.39
since from Beer's law, A= ebC
concentration =A/eb= 1.39/92.4 =0.015 moles/L
since X+ and Y- are aquesous, Z+=0.015moles/L
let x= drop in concentration of X+ to reach equilibrium
at Equilibrium. [X+]= 0.0333-x, [Y]= 0.1333-x
[Z+]=x= 0.015
C)
hence [X+]= 0.0333-0.015
=0.0183M and
[Y-]= 0.1333-0.015
= 0.1183M
D)
K= [Z]/ [X+] [Y-]
= 0.015/(0.1183*0.0183)
K= 6.93
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