The molar absorptivities for methyl red in its acid form are 44640 M^-1cm^-1 @ 5

Question

The molar absorptivities for methyl red in its acid form are 44640 M^-1cm^-1 @ 524 nm and 3860 M^-1cm^-1 @ 425nm and for methyl red in its base form are 740 M^-1cm^-1 @ 524nm and 18530 M^-1cm^-1 @ 425nm. The acid dissociation constant (Ka) of methyl red of a solution of pH=5.61 is 1.37*10^-5. Using the molar absorptivities for methyl red in its acid form given above, calculate the percentage of methyl red acid form in a solution with absorbances of 0.207 at 524 nm and 0.455 at 425 nm. Book answer is 15%. I don't understand how the book got 15% please help, clear explanation for 5 stars!!!

Explanation / Answer

let C1 be acid form conc , C2 is base conc , total absorption = sum of individual absorptions

A1+A2 = A = 0.207 = ( C1(44640) + C2(740) , (since A= eCl , l =1),

A = 0.455 = (C1(3860) + C2(18530) , solving these two we get C1 = 0.0000042447,

C2 = 0.00002367, C1 = ( 4.244x10^ -6) , C2 = 2.367 x10^ -5 ,

% in acid form = (4.2447 x10^ -6)/(2.367 x10^ -5) x100 =17.74 %

there might be error in calculation , but procedure is same ,

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