Question

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.

Answer 1

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:

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

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