Question

The absorbance of a solution of an analyte having 75% transmittance would be equal to (Given log 5= 0.6990 and log 3= 0.4771):

• A. 0.75
• B. 0.25
• C. 0.125
• D. 0.0625

Hint:

Alternatively, use the fractional transmittance formula: \(A = -\log_{10}(T) = -\log_{10}(0.75) = \log_{10}(4/3) = \log_{10}(4) - \log_{10}(3)\).
Since \(\log_{10}(2) = 1 - \log_{10}(5) = 1 - 0.699 = 0.301\), we have \(\log_{10}(4) = 2 \times 0.301 = 0.602\).
Thus, \(A = 0.602 - 0.4771 = 0.1249 \approx 0.125\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Absorbance (\(A\)) is related to percent transmittance (\(\%T\)) by a logarithmic relationship defined by the Beer-Lambert law.
Key Formula or Approach:
The equation relating absorbance and transmittance is: \[ A = 2 - \log_{10}(\%T) \]

Step 2: Detailed Explanation:
The percent transmittance is given as: \[ \%T = 75 \] Using the logarithmic equation: \[ A = 2 - \log_{10}(75) \] We can express 75 as a product of prime factors: \[ 75 = 3 \times 25 = 3 \times 5^2 \] Using log rules to expand: \[ \log_{10}(75) = \log_{10}(3) + \log_{10}(5^2) \] \[ \log_{10}(75) = \log_{10}(3) + 2\log_{10}(5) \] Substitute the given values \(\log_{10}(3) = 0.4771\) and \(\log_{10}(5) = 0.6990\): \[ \log_{10}(75) = 0.4771 + 2(0.6990) \] \[ \log_{10}(75) = 0.4771 + 1.3980 = 1.8751 \] Now, calculate the absorbance: \[ A = 2 - 1.8751 = 0.1249 \approx 0.125 \]

Step 3: Final Answer:
The absorbance value is approximately 0.125, which corresponds to option (C).