The extinction coefficient for NADH is 6220 M-1 cm-1 at 340 nm

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The extinction coefficient for NADH is 6220 M-1 cm-1 at 340 nm. Calculate the following:  The absorbance of a 2.2×10-5 M sample in a 1 cm cuvette at 340 nm. The absorbance of a 2.2×10-5 M sample in a 1 mm cuvette at 340 nm. The absorbance of a 1 mM sample in a 1 cm cuvette at 340 nm.
The extinction coefficient for NADH is 6220 M-1 cm-1 at 340 nm. Calculate the following:

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The absorbance of a 2.2×10-5 M sample in a 1 cm cuvette at 340 nm.
The absorbance of a 2.2×10-5 M sample in a 1 mm cuvette at 340 nm.
The absorbance of a 1 mM sample in a 1 cm cuvette at 340 nm.

The Correct Answer and Explanation is :

To calculate the absorbance of a solution using the Beer-Lambert law, we use the equation:

[
A = \varepsilon \cdot c \cdot l
]

Where:

  • ( A ) is the absorbance,
  • ( \varepsilon ) is the molar extinction coefficient (6220 M(^{-1}) cm(^{-1})),
  • ( c ) is the concentration of the sample (in molarity, M),
  • ( l ) is the path length of the cuvette (in cm).

1. Absorbance of a 2.2 × 10(^{-5}) M sample in a 1 cm cuvette at 340 nm:

Given:

  • ( \varepsilon = 6220 \, \text{M}^{-1}\text{cm}^{-1} ),
  • ( c = 2.2 \times 10^{-5} \, \text{M} ),
  • ( l = 1 \, \text{cm} ).

Substitute into the Beer-Lambert law:

[
A = 6220 \, \text{M}^{-1}\text{cm}^{-1} \times 2.2 \times 10^{-5} \, \text{M} \times 1 \, \text{cm}
]

[
A = 0.137 \, \text{(unitless, absorbance)}
]

2. Absorbance of a 2.2 × 10(^{-5}) M sample in a 1 mm cuvette at 340 nm:

In this case, the cuvette path length is 1 mm, which is equal to 0.1 cm.

Given:

  • ( \varepsilon = 6220 \, \text{M}^{-1}\text{cm}^{-1} ),
  • ( c = 2.2 \times 10^{-5} \, \text{M} ),
  • ( l = 0.1 \, \text{cm} ).

Substitute into the Beer-Lambert law:

[
A = 6220 \, \text{M}^{-1}\text{cm}^{-1} \times 2.2 \times 10^{-5} \, \text{M} \times 0.1 \, \text{cm}
]

[
A = 0.0137 \, \text{(unitless, absorbance)}
]

3. Absorbance of a 1 mM sample in a 1 cm cuvette at 340 nm:

For this calculation, convert 1 mM to molarity: 1 mM = 0.001 M.

Given:

  • ( \varepsilon = 6220 \, \text{M}^{-1}\text{cm}^{-1} ),
  • ( c = 0.001 \, \text{M} ),
  • ( l = 1 \, \text{cm} ).

Substitute into the Beer-Lambert law:

[
A = 6220 \, \text{M}^{-1}\text{cm}^{-1} \times 0.001 \, \text{M} \times 1 \, \text{cm}
]

[
A = 6.22 \, \text{(unitless, absorbance)}
]

Explanation:

The Beer-Lambert law describes the relationship between the absorbance of a sample, its concentration, and the path length of the cuvette. The higher the concentration and the longer the path length, the greater the absorbance. In the case of the first two calculations, we see a clear relationship between cuvette path length and absorbance. The shorter the path length (1 mm in the second case), the lower the absorbance, as seen in the values calculated. Similarly, for the third case, the higher concentration (1 mM compared to 2.2 × 10(^{-5}) M) leads to a much higher absorbance value. This demonstrates how absorbance can be used to determine concentration using known values for extinction coefficient and path length.

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