HIn < === > H+ + In- The following absorbance data were obtained for a 5.00 x 10
ID: 723710 • Letter: H
Question
HIn < === > H+ + In- The following absorbance data were obtained for a 5.00 x 10-4 M solution of HIn in 0.1 M NaOH and 0.1 HCl. Measurements were made at a wavelength of 485 nm and 625 nm with 1.00 cm cell.0.1 M NaOH A(485 nm)=0.052, A(625 nm)=0.823
0.1 M HCl A(485 nm)=0.454, A(625 nm)=0.176
In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially
all in the form of HIn.
A 25.00 mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a
standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the
base were added to a second 25.00 mL aliquot of the acid, which contained a small amount of the
indicator under consideration, the absorbance was found to be 0.306 at 485 nm and 0.555 at 625 nm
(1.00 cm cells). Calculate the pH of the solution and Ka for the weak acid. What would be the absorbance of a solution at 485 and 625 nm (1.25 cm cells) that was 2.00 x 10-4 M
in the indicator and was buffered to a pH of 6.000?
Please solve step-by-step and explain (i couldn't break the question because the first question is needed for the second one.) i hope you understand me. i appreciate your time and help
Explanation / Answer
ph=7.6(approx)
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