An acid-base indicator, HIn, dissociates according to the following reaction in

Question

An acid-base indicator, HIn, dissociates according to the following reaction in an aqueous solution.The protonated form of the indicator, HIn, has a molar absorptivity of 3621 M–1·cm–1 and the deprotonated form, In–, has a molar absorptivity of 2.480 × 104 M–1·cm–1 at 440 nm. The pH of a solution containing a mixture of HIn and In– is adjusted to 6.07. The total concentration of HIn and In– is 0.000171 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.855. Calculate pKa for HIn.

Explanation / Answer

For HIn, E(HIn) = 3621 M-1cm-1 = A / lC(HIn)

=> 3621 M-1cm-1 = 0.855 / 1cm x C(HIn)

=> C(HIn) = 0.855 / 1cm x 3621 M-1cm-1

For In-, E(In-) = 24800 M-1cm-1 = A / lC(In-)

=> 24800 M-1cm-1 = 0.855 / 1cm x C(In-)

=> C(In-) = 0.855 / 1cm x 24800 M-1cm-1

Applying Hendersen equation

pH = pKa + log[C(IN-) / C(HIn)]

=> 6.07 = pKa + log(3621 / 24800)

=> pka = 6.90

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