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Concentration Determination

  • Page ID
    280652

  • Name: _________________________

    1) Species of arsenic found in drinking water include AsO33- (arsenite), AsO43- (arsenate), (CH3)AsO2- (dimethylarsinate) and (CH3)AsO32- (methylarsonate). Pure water containing no arsenic was spiked with 0.40 μg arsenate/L.  Seven replicate determinations gave an average of  0.3829 and a standard deviation of  0.02138.  Find the mean percent recovery of the spike.

     

     

     

     

     

    2) If the least squares equation for a calibration curve is y=2.6x + 0.51, what is the concentration limit of quantitation if the mean and standard deviations of 10 measurements were 10.25 and 0.85, respectively?

     

     

     

     

    3) An unknown sample of Cu2+ gave an absorbance of 0.262 in atomic absorption analysis.  Then 1.00mL of solution containing 100.0 ppm Cu2+ was mixed with 95.0 mL unknown, and the mixture was diluted to 100.0 mL in a volumetric flask.  The absorbance of the new solution was 0.500.  What is the concentration of Cu2+ in the unknown?

     

     

     

     

     

     

     

     


     


    Name: _________________________

    4) Given the following data where an increasing amount of Sr was added to the same vial of unknown, determine the amount of strontium in ppm in the unknown.  The equation of the best fit line is y=3.136x + 27.36 for a plot of [Sf] vs. signal. 

     

     

     

     

     

    5) For a destructive AA analysis of Na+ in a sample, the following flasks were prepared, in which each flask contained 25.00 mL of unknown, varying additions of 2.640 M NaCl standard, and a total volume of 50.0 mL.  If a plot of emission signal vs. volume of standard added gives a best fit line equation of  y=2.36x + 3.12, what is the concentration of Na+ in the unknown?

    Flask

    Volume of STD added (mL)

    Na+ atomic emission signal (mV)

    1

    0

    3.13

    2

    1.000

    5.40

    3

    2.000

    7.89

    4

    3.000

    10.30

    5

    4.000

    12.48

     

     

     

     

     

    6) A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3,473 and 10,222, respectively in a chromatographic analysis.  The 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixture was diluted to 10.0 mL.  This solution gave peak areas of 5,428 and 4,431 for X and S, respectively.  Calculate the concentration of X in the original unknown.

     

     

     

     

     

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