Question

The molar absorptivity of the bi(iii) thiourea complex is 5.2x101cm-1m-1 at 470nm. What concentration (in mm) of the complex would be needed to produce a solution with a 6.85?

Answer 1

To achieve an absorbance of 6.85 for the bi(iii) thiourea complex at 470 nm, a concentration of 131.7 mM is required, calculated using Beer's Law.

Step-by-step explanation:

The student has asked how to calculate the concentration of a bi(iii) thiourea complex in millimolar (mM) units given that the molar absorptivity is 5.2x101 cm-1mol-1 at 470 nm and aiming for an absorbance value of 6.85. The equation that links absorbance (A), molar absorptivity (ε), concentration (c), and path length (l) is given by Beer's Law, which states A = εcl. Assuming that the path length (l) is standard at 1 cm, the equation simplifies to A = εc. To find the concentration, we rearrange this equation to c = A/ε. Thus, for our values, c = 6.85 / (5.2x101), which gives us the concentration in mol/L. To convert this to millimolar (mM), we multiply by 1000.

Calculation:

c = 6.85 / (5.2x101)

c = 0.1317 mol/L

c = 131.7 mM

Therefore, the required concentration of the bi(iii) thiourea complex is 131.7 mM to achieve an absorbance of 6.85 at the specified conditions.