Detection limit. In spectrophotometry, we measure the concentration of analyte by its absorbance of light. A lowconcentration
sample was prepared, and nine replicate measurements
gave absorbances of 0.004 7, 0.005 4, 0.006 2,
0.006 0, 0.004 6, 0.005 6, 0.005 2, 0.004 4, and 0.005 8 in a
1.000-cm cell. Nine reagent blanks gave values of 0.000 6,
0.001 2, 0.002 2, 0.000 5, 0.001 6, 0.000 8, 0.001 7, 0.001 0,
and 0.001 1.
(a) Find the absorbance detection limit with Equation 5-2.
(b) The calibration curve is a graph of absorbance versus
concentration in a cell with a pathlength of 1.000 em. Absorbance
is a dimensionless quantity. The slope of the calibration
curve is m = 2.24 X 104 M-1• Find the concentration detection
limit with Equation 5-4.

answer: - first one is the correct answer. .

explanations: - potassium sulfide is and cobalt(ii) chloride is .

charge on k is +1, charge on s is -2, charge on co is +2 and charge on cl is -1.

in a double replacement reaction the ion exchange takes place. in general the reaction looks like:

if one of the product is insoluble as per solubility rules then a precipitate of it is formed.

when we mix the aqueous solutions of potassium sulfide and cobalt(ii) chloride then chloride ion goes with potassium ion and sulfide ion goes with cobalt ion and the balanced equation is written as:

k and cl has equal charges, similarly co and s has equal charges and hence no subscripts are used for the product side.

as per the solubility rules, cos is insoluble and so it's precipitate is formed.

looking at the given choices, first choice is correct.