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maksim [4K]
3 years ago
5

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.
Chemistry
1 answer:
enyata [817]3 years ago
7 0

Answer:

Ka=0.258

Explanation:

To calculate the dissosiation factor of the HIn first we need to determine the concentration of ions in the solution. To do that we use the Lambert-Beer's law:

A=a*b*c

Where:

A is the absorbance

a is the absorptivity

b is the length of the cuvette

c is the concentration

0.818=a*1cm*0.000127M

a=6440.9 cm^{-1}*M^{-1}

With this absorptivity we can calculate the concentration of HIn and In:

x_{HIn]+x_{In}=1

x_{HIn]=1-x_{In}

a=x_{In}*20060cm^{-1}*M^{-1}+ x_{HIn]*2929cm^{-1}*M^{-1}

6440.9 cm^{-1}*M^{-1}=x_{In}*20060cm^{-1}*M^{-1}+ (1-x_{In})*2929cm^{-1}*M^{-1}

6440.9 cm^{-1}*M^{-1}=x_{In}*20060cm^{-1}*M^{-1}+ 2929cm^{-1}*M^{-1} -x_{In}*2929cm^{-1}*M^{-1}

x_{In}*17131cm^{-1}*M^{-1}=3511cm^{-1}*M^{-1}

x_{In}=0.205

x_{HIn]=1-0.205=0.795

For the Ka:

Ka=\frac{[In]}{[HIn]}

Ka=\frac{0.205}{0.795}=0.258

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