The protonated form of the indicator, HIn, has a molar absorptivity of 2051 M–1·
ID: 1058929 • Letter: T
Question
The protonated form of the indicator, HIn, has a molar absorptivity of 2051 M–1·cm–1 and the deprotonated form, In–, has a molar absorptivity of 14050 M–1·cm–1 at 440 nm. The pH of a solution containing a mixture of HIn and In– is adjusted to 6.18. The total concentration of HIn and In– is 0.000146 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.805. Calculate pKa for HIn.
EA Sapling Learning macmillan learning An acid-base indicator, Hln, dissociates according to the following reaction in an aqueous solution. HIna In (aq) H (a The protonated form of the indicator, Hln, has a molar absorptivity of 2051 M cm 1 and the deprotonated form, In has a molar absorptivity of 14050 M Cm 1 at 440 nm. The pH of a solution containing a mixture of HIn and In s adjusted to 6.18. The total concentration of HIn and In s 0.000146 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.805. Calculate pKa for HIn Number pK, HIMExplanation / Answer
We know ,
A=Ecb
0.805 = 2051xC1 x 1 + 14050 x C2 x 1 ------------ 1
0.000146 = C1 + C2 ------------- 2
Solve (1) and (2),
C1= 0.000104
C2 =0.0000421
pH =pKa + log(C2/C1)
pKa =6.18 -log(0.0000421 /0.000104 )
= 6.573
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