A solution was prepared by mixing 25.00 mL of 0.080 0 M aniline, 25.00 mL of 0.060

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A solution was prepared by mixing 25.00 mL of 0.080 0 M aniline, 25.00 mL of 0.060 0 M sulfanilic acid, and 1.00 mL of 1.23 × 10^-4 M HIn and then diluting to 100.0 mL. (HIn stands for protonated indicator.)

NH3

The absorbance measured at 550 nm in a 5.00-cm cell was 0.110. Find the concentrations of HIn and In- and pKa for HIn.

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Quantitative Chemical Analysis

ISBN: 978-1429218153

8th edition

Authors: Daniel C. Harris

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Question Posted: Apr 19, 2016 10:04 AM

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A solution was prepared by mixing 25.00 mL of 0.080 0 M aniline, 25.00 mL of 0.060

Question:

NH3 "0,S- NH, Sulfanilic acid Anilinium ion pк, — 3.232 pK, = 4.601 HIn H* + In = 2.45 x 10 M- cm 2.26 x 10' M-' cm" E325 = 4.39 x 10' M-'cm Ess0 = 1.53 X 10 M-'cm-! Esso

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The quantity of HIn is small compared to aniline and sulfanilic acid Calling ani...View the full answer

Quantitative Chemical Analysis

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