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Answer Happy • b. When two drops of concentrated HCI were added to the pH 7.13 solution, the indicator was converted to pure Hin form.
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b. When two drops of concentrated HCI were added to the pH 7.13 solution, the indicator was converted to pure Hin form.

Posted: Sun Jul 10, 2022 3:49 pm
by answerhappygod
B When Two Drops Of Concentrated Hci Were Added To The Ph 7 13 Solution The Indicator Was Converted To Pure Hin Form 1
B When Two Drops Of Concentrated Hci Were Added To The Ph 7 13 Solution The Indicator Was Converted To Pure Hin Form 1 (39.79 KiB) Viewed 63 times
b. When two drops of concentrated HCI were added to the pH 7.13 solution, the indicator was converted to pure Hin form. Its measured absorbance, At was 0.003. When two drops of 50% NaOH were added to the pH 8.56 solution, the result was the pure in form. The measured absorbance of this solution, An, was 0.712. Calculate 7.13 7.39 7.79 8.21 8.56 for each of the five solutions. Algebraically determine the pK, for each solution, using Equation 5, then determine the average pK,. Calculate K, from the average pk, pH K₁= A 0.144 0.23 0.4 (A-At-) (AHin - A) K y-intercept = average pK, and log (A-A) (AHIn-A) 4.03 0.55 0.63 2.00 X10-8 2.12 0.786 0.296 0.131 c. Using the graph paper provided, graph pH versus [(A-An) (AHin - A)] (A-A) [(AHIN-A) log [(A-Ain (AH-A) log 0.605 0.326 -0.105 -0.529 -0.883 average pk, = pk, 7.74 7.72 Determine the y-intercept, which should equal the average pK,. Calculate K, from the graphically- determined average pk,. 761 2.68 7.68 7.76