A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standa

Question

A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 mu g S/mL, and diluting to 50.0 mL. The measured signal quotient was (signal due to X/signal due to S) = 1.490/1.000. (a) In a separate experiment it was found that, for equal concentrations of X and S, the signal quotient was (signal due to X/signal due to S) = 0.960/1.000. Calculate the concentration of X in the unknown. (b) Answer the same question if in a separate experiment it was found that for the concentration of X equal to 3.42 times the concentration of S, the signal quotient was (signal due to X/signal due to S) = 0.930/1.000.

Explanation / Answer

From the given data,

Concentration of standard in 50 ml = 8.24 ug x 5/50 = 0.824 ug

Signal ratio (X/S) = 1.490/1.000

(a) For [X] = [S]

signal ratio (X/S) = 0.960/1.000

So,

concentration of X in the unknown = 1.49/[(0.96)(1/0.824)] = 1.28 ug in 50 ml

Concentration of X in original solution = 1.28 x 50/10 = 6.4 ug

(b) when [X] = 3.42[S]

concentration of X in the unknown = 1.49/[(0.93/3.42)(1/0.824)] = 4.514 ug in 50 ml

Concentration of X in original solution = 1.28 x 50/10 = 22.57 ug

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