A student performed a lab similar to this one in an attempt to find Keq for the
ID: 711840 • Letter: A
Question
A student performed a lab similar to this one in an attempt to find Keq for the equilibrium system X+(aq) + Y-(aq) Z (blue, aq). 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, , is known to be 92.4 L•mol-1.
1.) Calculate the initial molarity of each reactant added to the test tube at the moment of mixing, [X+]init and [Y-]init. (HINT: these are dilution calculations.
2.)Calculate the value of A, then find [Z]equil.
3.)Set up an I.C.E. table to find [X+]equil and [Y-]equil.
4.)Find the value of Kc for this equilibrium system.
Explanation / Answer
Given data,
1)
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
=0.0333M and
Y=0.002/0.015
= 0.1333M
2)
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
3)
hence [X+]= 0.0333-0.015
= 0.0183M and
[Y-]= 0.1333-0.015
= 0.1183M
4)
K= [Z]/ [X+] [Y-]
= 0.015/(0.1183*0.0183)
= 6.93
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