The protonated form of the indicator, HIn, has a molar absorptivity of 2929 M–1·

Question

The protonated form of the indicator, HIn, has a molar absorptivity of 2929 M–1·cm–1 and the deprotonated form, In–, has a molar absorptivity of 20060 M–1·cm–1 at 440 nm. The pH of a solution containing a mixture of HIn and In– is adjusted to 6.12. The total concentration of HIn and In– is 0.000127 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.818. Calculate pKa for HIn.

An acid-base indicator, Hln, dissociates according to the following reaction in an aqueous solution. HInlag) In (aq) H (aq) The protonated form of the indicator, Hln, has a molar absorptivity of 2929 M cm 1 and the deprotonated form, In has a molar absorptivity of 20060 M-1. cm 1 at 440 nm. The pH of a solution containing a mixture of Hin and In s adjusted to 6.12. The total concentration of HIn and In s 0.000127 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.818. Calculate pKa for HIn. Number

Explanation / Answer

Part A:

Let,       [HIn] = X M       and       [In-] = Y M                     [M = molar]

            So, X + Y = 0.000127 M             -- equation 1

Beer-Lambert’s Law, A = e C L             - equation 1,               where,

                                                A = Absorbance

                                                e = molar absorptivity at specified wavelength

                                                L = path length (in cm)

                                                C = Molar concentration of the solute

Absorbance of Hin, A = eCL = (2929M-1cm-1) x X M x 1 cm

                        Or, A (HIn) = 2929 X

Absorbance of in, A = eCL = (20060 M-1cm-1) x Y M x 1 cm

                        Or, A (In) = 20060 Y

Give, total absorbance at 440 nm = 0.8

So, 2929 X + 20060 Y= 0.818     ---equation 3

Comparing equation 1 minus (equation 3 / 2929)

Or,

            X          +          Y          =          0.000127          

       -     X          + 6.84875384 Y =          0.000279

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Or,                    - 5.84875384 Y =         - 0.000152

Or, Y = 0.000025988442

Hence [In] = Y = 0.000025988442 M

Putting the value of Y in equation 1,

            X = 0.000127 M - 0.000025988442 M = 0.000101011558 M

Hence, [HIn] = X = 0.000101011558 M

Now, using Henderson- Hasselbalch equation,

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

            or, 6.12 = pKa + log (0.000025988442 M / 0.000101011558 M)

            or, 6.12 = pKa + (- 0.5895)

            or, pKa = 6.12 + 0.59 = 6.71

Hence, pKa of the solution = 6.71

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