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

Question

An acid-base indicator. Hln, dissociates according to the following reaction in an aqueous solution Hln(aq) ln^- (aq) + H^+ (aq) The protonated form of the indicator, Hln, has a molar absorptivity of 2607 M^-1 cm^-1 and the deprotonated form. In^-, has a molar absorptivity of 17860 M^-1 cm^-1 at 440 nm The pH of a solution containing a mixture of Hln and ln^- is adjusted to 5.78. The total concentration of Hln and ln^- is 0.000198 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.752. Calculate pK_a for Hln.

Explanation / Answer

For HIn, E(HIn) = 2607 M-1cm-1 = A / lC(HIn)

=> 2607 M-1cm-1 = 0.752 / 1cm x C(HIn)

=> C(HIn) = 0.752 / 1cm x 2607 M-1cm-1

For In-, E(In-) = 17860 M-1cm-1 = A / lC(In-)

=> 17860 M-1cm-1 = 0.752 / 1cm x C(In-)

=> C(In-) = 0.752 / 1cm x 17860 M-1cm-1

Applying Hendersen equation

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

=> 5.78 = pKa + log(2607 / 17860)

=> pka = 6.616 (answer)

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