Question

Is your unknown more, less, or the same concentration as the standard? Explain using %T and/or Absorbance.

LAB DATA Enter Lab Data before leaving lab 1 Click the Next Problem button. 2. Enter data in the 1" text box and click the Submit Data button. text boxes. Enter data in the dlick the Submit Data button. Concentration (M) of starting solution 0024 1st Reading 2nd Reading % Transmittance. %T . ( Blank ) 97.2 %T ( 15-mL 0.024 M Cu+7 mL conc NH, diluted with distilled H,O 100 mL total volume) %T (5-mL 0,024 M Cu? 7 mL conc NH, diluted with distilled H.O to 100 mL total volume) Unknown Identification Number 21 %T ( 15-ml Unknown Cut solution +7 mL conc NH, diluted with distilled H-0 to 100 m total volume) 37.2 % T (5-mL Unknown Cu solution + 7 mL conc NH, diluted with distilled H,0 to 100 mL total volume) 68 2 LAB CALCULATIONS %T ( Trial 1) Sample %T (Trial 2) Absorbance Average % T Try 1:97. Blank 97.25 4 Try 1:58.45 Try 1:0220 15-ml Dilution 58.9 44 Try 030 Ratio M/A (use 0.024 M as the given M) for the 15-mL Dilution 4 %T ( 1st Reading) % T (2nd Reading) Sample Absorbance Average % T Try 1:81.25 Try 1:0.0773. 5-mL Dilution 81.2 81.3 Try 1:03104 Ratio M/A (use 0.024 M as the Eiven M) for the S-ml, Dilution Unknown Identification Number % T ( 1st Reading) Sample %T (2nd Reading) Absorbance Average % T Try 137.1 Try 1:0417 Unknown (1" Dilution) 37 37,2 4 44 Try 1:68.1 Try 1:0154 Unknown (2ed Dilution) 68.2

Answer 1

From the Beer-Lambert's law, the relation between concentration and absorbance can be given as follows.

A = $\varepsilon$CL

A = absorbance;    C = Molar concentration; L = length of Cell and  $\varepsilon$ = Molar extinction coefficient.

Here, L and $\varepsilon$ are non variable (fixed) so the product L X $\varepsilon$ is a constant

So, the only variable that alter that defines the Absorbance is concentration.

A $\alpha$ C

As concetration increases absorbance increases and vice versa.

From the camparsion of Absorbances of Know and Unknown

Absorbance

15 mL dilution of Known 0.220

15 mL dilution of Unknown 0.417

It clear from the observation that the absobance of unknown (0.417) > absorbance of Known (0.220)

Absorbance

5 mL dilution of Known 0.0773

5 mL dilution of Unknown 0.154

It clear from the observation that the absobance of unknown (0.154) > absorbance of Known (0.0773)

From both observation, it clearly indicates that the concentration of unknown is more compared to known.

For 15 mL dilution

AbsorbanceofUnknownXConcentrationof K nown Concentrationof Unknown AbsorbanceofKnown

Concentration of Unknown = 0.417 X 0.024 M / 0.220 = 0.0455 M

For 5 mL dilution

Concentration of Unknown = 0.154 X 0.024 M / 0.0773 = 0.0478 M