Determine the molar absorptivity of the acidic form of bromocresol green at 445

ID: 709331 • Letter: D

Question

Determine the molar absorptivity of the acidic form of bromocresol green at 445 nm and at 615 nm.

Stock solution of bromoscresol green contains 80.0 mg/L

Molar mass of bromocresol green is 698.0 g/mol

*This was done in a lab so I'll put down some information if needed:

10mL of bromocresol green was mixed with 40mL of .4M HCl

10mL of bromocresol green was mixed with 40mL of .4M NaOH

Here are the absorbances at 445nm:

.4M HCl plus the bromocresol green= .382A

.4M NaOH plus the bromocresol green= .045A

Here are the absorbances at 615nm:

.4M HCl plus the bromocresol green= .003A

.4M NaOH plus the bromocresol green= .872A

10 ml bromocresol contains = (10/1000) x 80/L = 0.8 mg = 8x10^ -4 gm,

moles = 8x10^ -4 /689 = 1.16 x10^ -6

when 10 ml bromocresol mixed with 40 ml HCl , total vol = 50 ml =0.05 liter

cocn = ( 1.16x10^ -6 )/(0.05) = 2.3 x10^ -5 M

at 445 nm

when HCl added A= 0.382 = e (2.3x10^ -5) , e = 16608.7 M-1cm-1

when NaOH added A= 0.045= (2.3x10^ -5) , e = 1956.5 M-1cm-1,

at 615 nm

when HCl added A= 0.03= e(2.3x10^ -5) , e = 1304.35 M-1cm-1,

when NaOh is added A= 0.872 = e(2.3x10^ -5) , e = 37913 M-1cm-1

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