Question

You are quantitating benzene using isopropyl-benzene as an internal standard. During the first run, you add 100.0 microliters of 500. microgram/mL benzene and 100.0 microliters of 500. microgram/mL isopropylbenzene to a vial and dilute to 1.000mL to prepare a solution that has a final concentration of 50.0 micrograms/mL of benzene and isopropylbenzene. The peak area measured by GC-MS corresponds to 1225 and 1775, respectively.

Taking an unknown sample of a solution containing benzene with a volume of 100.0 microliters, you spike it with 100.0 microliters of 500. microgram/microliter solution of isopropylbenzene and dilute to the 1.000 mL mark. You measure a peak area of 1550 for benzene and 1800 for isopropyl benzene.

The information is organized in the table below with the appropriate dilution factors applied for what is in the sample vial.

	Standard	Spiked Sample
[Benzene], micrograms/mL	50.0	Unknown
Benzene peak area	1225	1550
[Isopropylbenzene], micrograms/mL	50.0	50.0
Isopropylbenzene peak area	1775	1800

What is the concentration of Benzene in the unknown sample, in units of micrograms/mL?

A series of volumetric flasks are being prepared for the analysis of mercury in the drinking water supply using the method of standard addition. To prepare the flasks, 10 mL of drinking water and a varying amount of standard are added to each flask, and the flasks are brought to the 25.00mL mark on the volumetric flask.

The concentration of mercury standard used was 20.0 microgram/mL.

Please use the data tabulated below to provide the concentration of mercury in the drinking water, in units of micrograms/mL.

It is recommended to use Excel or similar software

mL Mercury Standard Added	Signal
0.00	19.65
1.00	30.35
2.00	40.28
3.00	49.78
4.00	60.44
5.00	70.06

Please Show All Steps

Answer 1

1) Let the area be related to the concentration as

A = K*(concentration)

where K = constant of proportionality.

Determine the value of K for isopropylbenzene (call this K’) and check the authenticity of the linear relationship.

We have for isopropylbenzene,

1775 = K’*(50.0 µg/mL)

=====> K’ = (1775)/(50.0 µg/mL)

=====> K’ = 35.5 mL/µg

Again,

1800 = K’*(50.0 µg/mL)

=====> K’ = (1800)/(50.0 µg/mL)

=====> K’ = 36.0 mL/µg

The value of K’ is similar within the limits of experimental error and thus, the linear relationship is valid.

Let the constant for benzene be K”. Therefore,

1225 = K”*(50.0 µg/mL)

=====> K” = (1225)/(50.0 µg/mL)

=====> K” = 24.5 mL/µg

Determine the concentration of benzene in the unknown sample (dilute) as below.

1550 = (24.5 mL/µg)*(concentration of benzene)

=====> concentration of benzene = (1550)/(24.5 mL/µg)

=====> concentration of benzene = 63.2653 µg/mL.

The above denotes the concentration of benzene in the diluted portion of the unknown sample.

The diluted portion was prepared by diluting 100.0 µL of the unknown sample to a total volume of 1.000 mL. The dilution factor = (final volume)/(volume taken)

= (1.000 mL)/(100.0 µL)

= (1.000 mL)*(1000 µL)/(1 mL)/(100.0 µL)

= 10.

Concentration of benzene in the unknown sample = (concentration in dilute portion)*(dilution factor)

= (62.2653 µg/mL)*(10)

= 622.653 µg/mL ≈ 622.6 µg/mL (ans).

2) The concentration of the added mercury standard in the final solutions can be obtained by using the dilution equation.

C₁*V₁ = C₂*V₂

where C₁ = concentration of the stock solution = 20.0 µg/mL; V₁ = volume of stock solution added and V₂ = final volume of the test solutions = 25.00 mL.

Take solution 2 as an example. We have V₁ = 1.00 mL.

Plug in values and get

(20.0 µg/mL)*(1.00 mL) = C₂*(25.00 mL)

======> C₂ = (20.0 µg/mL)*(1.00 mL)/(25.00 mL)

======> C₂ = 0.80 µg/mL.

Prepare the table.

Trail	Volume of mercury solution added (mL)	Concentration of mercury solution added (µg/mL)	Signal
1	0.00	0.00	19.65
2	1.00	0.80	30.35
3	2.00	1.60	40.28
4	3.00	2.40	49.78
5	4.00	3.20	60.44
6	5.00	4.00	70.06

Plot absorbance vs concentration as below.

Plot of absorbance vs concentration

The concentration of mercury in the unknown sample is obtained by setting y = 0 in the regression equation. Put y = 0 and obtain

0 = 12.565x + 19.963

=====> -12.565x = 19.963

=====> x = -19.963/12.565

=====> x = -1.5887

Ignore the negative sign (concentration cannot be negative) and report the concentration as 1.5887 µg/mL.

The unknown solution was prepared by taking 10.00 mL of the solution and diluting to 25.00 mL; therefore, the concentration of mercury in the unknown solution

= (1.5887 µg/mL)*(25.00 mL)/(10.00 mL)

= 3.97175 µg/mL

≈ 4.0 µg/mL (ans).