Question

The absorbance of a cationic iron(II) sample solution was measured in a spectrophotometer, but the instrument returned an error because the absorbance was too high. The sample was then diluted by using a pipette to take 100.0 μL of the sample and injecting it into a cuvette already containing 2.00 mL of water (total volume is 2.00 mL + 100.0 μL). The absorbance value of the diluted solution corresponded to a concentration of 6.41×10−6 M . What was the concentration of the original solution? Express the concentration to three significant figures with the appropriate units.

Answer 1

Answer:

$1.35\times 10^{-4} M$ was the concentration of the original solution.

Explanation:

$M_1V_1=M_2V_2$  (Dilution)

where,

$M_1\text{ and }V_1$ are molarity and volume of non diluted sample.

$M_2\text{ and }V_2$ are molarity and volume of diluted sample.

$M_1=?$

$V_1=100.0 \mu L=0.1 mL$

(1μL=0.001  mL)

$M_2=6.41\times 10^{-6} M$

$V_2=2.00 mL + 100.0 \mu L=2.00 mL+0.1 mL=2.1 mL$

Substituting all values :

$M_1\times 0.1 mL=6.41\times 10^{-6} M\times 2.1 mL$

$M_1=\frac{6.41\times 10^{-6} M\times 2.1 mL}{0.1 mL}=1.35\times 10^{-4} M$

$1.35\times 10^{-4} M$ was the concentration of the original solution.