A solution containing 3.47 mM X (analyte) and 1.72 MS standard) gave peak areas

Question

A solution containing 3.47 mM X (analyte) and 1.72 MS standard) gave peak areas of 3473 and 10 222, respectively, in a chromatographic analysis. mM Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixture was diluted to 10.00 mL This solution gave peak areas of 5 428 and 4 431 for X and S, respectively. Calculate the response factor for the analyte. Find the concentration of S (mM) in the 10.0 mL of mixed solution. Find the concentration of x (mM) in the 10.0 mL of mixed solution. Find the concentration of X in the original unknown solution.

Explanation / Answer

a) response factor for X=peak area/concentration=RFX=3473/3.47mM=1000.864

b)CS=8.47mM

vs=1ml

Csfinal=?

total volume of mixture=Vf=10ml

using eqn,Cs*Vs=Csf*Vf

Csf=Cs*Vs/Vf=8.47mM*1ml/10ml=0.847mM

c)

RFS=response factor for S=peak area/concentration=10222/1.72*1000mM=5.943

Relative response factor=RRF=RFX/RFS=1000.864/5.943=168.410

So,

RRF=RFX/RFS=(peak area X/concentration X)/(peak area S/concentrationS)

or

RRF=(peak area X* Cs/Cx*peak area S)

or, CX=(peak area X* Cs/ peak area S)*1/RRF=(5428*0.847mM/4431 )*1/168.410=0.00616 mM

d)CX=unknown

Vx=5 ml

CXfinal=?

total volume of mixture=VX=10ml

using eqn,CX*VX=CXf*Vf

CXf=CX*VX/Vf=CX mM*5ml/10ml=Cx/2

or ,CXf=Cx/2.

or Cx=2*CXf=2*0.00616mM=0.0123mM

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