Sapling Learning macmillan learning An acid-base indicator, Hln, dissociates acc

Question

Sapling Learning macmillan learning An acid-base indicator, Hln, dissociates according to the following reaction in an aqueous solution aHIn HInlaq)n)(ay) The protonated form of the indicator, Hln, has a molar absorptivity of 2055 M1 cm 1and the deprotonated form, In-, has a molar absorptivity of 14080 M-1 cm-1 at 440 nm. The pH of a solution containing a mixture of HIn and In is adjusted to 6.14. The total concentration of Hln and In is 0.000196 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.879. Calculate pKa for HIn. Number HIn

Explanation / Answer

Ans. #Step 1: Let, at pH 6.14,       [HIn] = X molar         , and [In-] = Y molar.

So,

            X M + Y M = 0.000196 M

            Hence, X + Y = 0.000196               - equation 1

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

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M-1cm-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

The total abs of the solution is due to presence of both HIn and In-.

So,

            Abs of soln. = Abs of HIn + Abs of In-

            Or, 0.879 = (2055 M-1 cm-1 x X M x 1.0 cm) + (14080 M-1 cm-1 x Y M x 1.0 cm)

            Or, 0.879 = 2055X + 14080 Y

            Hence, 2055X + 14080 Y = 0.879                        - equation 2

# Comparing (equation 1 x 2055) – Equation 2

                 2055X + 2055 Y = 0.4028

            (-) 2055X + 14080 Y = 0.879         

            -------------------------------------

                                    -12025Y = -0.4762

                        Or, Y = 0.4762 / 12025 = 3.9602 x 10-5

Therefore, [In-] = Y molar = 3.9602 x 10-5 M

# Putting the value of Y in equation 1-

            X = 0.000196 M - 3.9602 x 10-5 M = 1.5640 x 10-4 M

# Step 2: Given- pH = 6.14

So, equilibrium [H3O+] = 10-pH = 10-6.14 = 7.2444 x 10-7

Now,

            Ka = [In-] [H3O+] / [HIn]                   - all equilibrium conc. at pH 6.14

            Or, Ka = (3.9602 x 10-5 x 7.2444 x 10-7) / (1.5640 x 10-4) = 1.8344 x 10-7

# pKa = -log Ka = -log (1.8344 x 10-7) = 6.7365

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