A #2.4# x #10^-4# #M# solution of a copper complex has transmittance of #58 %# when measured in a cell of path length #1.5# #cm#. Calculate the transmittance of solution if: (a) the concentration is doubled, (b) concentration is made half?

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Answer 1 · 2016-06-05T16:58:12

#(a)" "27%#

#(b)" "72.1%#

The Beer - Lambert Law states:

#A=epsiloncl#

#A# is the absorbance as measured by the spectrometer

#c# is the molar concentration

#l# is the path length of the cell in #"cm"#

The absorbance #A# can be written in terms of the transmittance #I# as:

#A=log[(I_0)/(I)]#

#I_0# is set to 100 so this becomes:

#A=log[100/I]=2-logI#

The first part of the question involves finding #epsilon#. Then this can be used to calculate #(a)# and #(b)#.

#2-logI=epsiloncl#

#:.epsilon=(2-logI)/(cl)=(2-1.763)/(2xx10^(-4)xx1.5)=790" ""mol"^(-1)."l"."cm"^(-1)#

#color(blue)((a))#

The concentration is doubled:

#2-logI=790xx4.8xx10^(-4)xx1.5=0.5688#

#:.logI=2-0.5688=1.4312#

#:.color(red)(I=27%)#

This seems a reasonable answer as you would expect less light to be transmitted if the solution is more concentrated.

#color(blue)((b))#

The concentration is halved:

#:.2-logI=790xx1.2xx10^(-4)xx1.5=0.1422#

#:.logI=2-0.1422=1.8578#

#color(red)(I=72.1%)#

Again, this seems a reasonable answer as we have diluted the solution so you would expect more light to pass through.