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

`Ka=0.258`

Step-by-step 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=([In])/([HIn])`

`Ka=(0.205)/(0.795)=0.258`