Seure 11/19/2017 1200AM 4o/o Toble Print akuistr Aailable From Due Date Points P

Question

Seure 11/19/2017 1200AM 4o/o Toble Print akuistr Aailable From Due Date Points Possible Grade CategoryC Sapling Learning An aaid-base Indicator, Hin, dissoaiaes acnrding to the following resction in an squcaus s Policies The protonated form of the indicator, Hin, has a molar absorptivity of 2737 M-1 om and the deprotonated fom, i, has a molar absorptivity of 18750 M"t crn-t at 440 nm The pH of a solution contaning a mixture of Hin and in is adjusted to 8.14. The total concentration of Hin and in is o0.000160 M. The absorbance of this solution was measured at 440 nm in a 1.00 om cuvette and was determined to be (0.829. Calculate pke for Hin. You can check your You can view solutie You have five atten You lose 10% ofthe answer in your que attempt at that ans D eTextbook °Help With This T O Web Help & Vide O Technical Suppor

Explanation / Answer

Let the concentration of HIn in the solution be x M. Since the total concentration of HIn and In- in the solution is 0.000160 M, we can express the concentration of In- as (0.000160 – x) M.

The absorbances are additive; hence we can write for the absorbance,

Absorbance = (molar absorptivity of HIn)*(concentration of HIn)*(path length of HIn solution) + (molar absorptivity of In-)*(concentration of In-)*(path length of In-)

====> 0.829 = (2737 M-1cm-1)*(x M)*(1.00 cm) + (18750 M-1cm-1)*[(0.000160 – x) M]*(1.00 cm)

====> 0.829 = 2737x + 18750*(0.000160 – x)

====> 0.829 = 2737x + 3.000 – 18750x

====> 0.829 – 3.000 = 2737x – 18750x

====> -2.171 = -16013x

====> x = 2.171/16013 = 0.0001356

Therefore, we have the concentration of HIn is 0.0001356 M and the concentration of In- is (0.000160 – 0.0001356) M = 2.44*10-5 M.

The acid dissociation constant is given as

Ka = [H+][In-]/[HIn]

Take logarithms on both sides and get

log (Ka) = log [H+] + log [In-]/[HIn]

====> - log (Ka) = -log [H+] – log [In-]/[HIn]

====> pKa = pH – log [In-]/[HIn] (we define pKa = -log Ka and pH = -log [H+])

Plug in values and get

pKa = 6.14 – log (2.44*10-5 M)/(0.0001356 M)

= 6.14 – log (0.17994) = 6.14 – (-0.74487) = 6.88487 6.88 (ans).

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