1. The acid HIn has an acid dissociation constant of $.20 x 10%. The following d
ID: 229126 • Letter: 1
Question
1. The acid HIn has an acid dissociation constant of $.20 x 10%. The following data are for a HIn solution of 7.50 x 10* M. Calculate the absorbance at 450 nm of a solution in which the total concentration of HIn is 7.5 x 103 M. The pH of the solution is 4.67 445 450 455 470 510 550 570 585 595 bsorbance at pH 1.0 0.657 0.658 0.656 0.614 0.353 0.119 0.068 0.044 0.032 0.068 0.076 0.085 0.116 0.223 0.324 0.352 0.360 0.361 0.35s 610 0.019 The following data have been determined previously. Calculate the concentrations of P and R in the mixture. 2. Compouncd Absorbance 365nm 0.150 0.684 0.71 Absorbance Concentration 0.642 0.088 0.604 x 10- M 2 x 10'M Unknown To determine the iron concentration in a California Burgundy wine, a 10-ml aliquot of the wine was added to each of four 50ml beakers. To the first beaker was added 10ml of water To 3. the second beaker was added 10ml of 4ppm Fe?: 10ml of 8ppm Fe was added to the third; and 10ml of 12ppm Fe* was added to the fourth. If the measured absorbances were 0.028, 0.045, 0.063, and 0.077, what was the concentration of iron in the original winc?Explanation / Answer
1) HIn <-->H+ +In-
At pH=4.67
pka(HIn/In-)=-logka=-log (5.20*10^-6)=5.3
pH=pka+log[In-]/[HIn] (Henderson-hasselbach equation)
4.67=5.3+log[In-]/[HIn]
or, [In-]/[HIn]=0.234/1
[In-]+[HIn]=7.5*10^-5M
[In-]=(0.234/(1+0.234))*(7.5*10^-5M)=1.422*10^-5M
[In-]=1.422*10^-5M=concentration of absorbing species
At 450nm,
if pH=1.0
Absorbance=e*l*C
where e=absorptivity of species
l=path length of light(1 cm)
plugging in the value of absorbance from table,
0.658=el*(7.5*10^-5M)
el=0.658/(7.5*10^-5M)=8773.333M^-1
So ,this el vaue is at wavelength=450nm
Using this value for above pH=4.67
A=(8773.333 M^-1)*(1.422*10^-5M)=0.125
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