ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
2nd Edition
ISBN: 9780393664034
Author: KARTY
Publisher: NORTON
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Question
Chapter 6, Problem 6.55P
Interpretation Introduction
Interpretation:
The relative acid strengths of the two acidic protons (pKa1 = -9, pKa2 = 2) in sulfuric acid are to be explained.
Concept introduction:
Stability of an anion depends mainly on the number of charges. Higher the charge, lower the stability. It also depends on delocalization of the charge. More the number of atoms on which the charge is delocalized, higher the stability of the anion.
Higher the stability of the anion formed on dissociation of the acidic proton, stronger the acid.
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Chapter 6 Solutions
ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
Ch. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10P
Ch. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.20PCh. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.23PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - Prob. 6.35PCh. 6 - Prob. 6.36PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38PCh. 6 - Prob. 6.39PCh. 6 - Prob. 6.40PCh. 6 - Prob. 6.41PCh. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - Prob. 6.52PCh. 6 - Prob. 6.53PCh. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Prob. 6.57PCh. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - Prob. 6.60PCh. 6 - Prob. 6.61PCh. 6 - Prob. 6.62PCh. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Prob. 6.72PCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. 6.83PCh. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - Prob. 6.86PCh. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.1YTCh. 6 - Prob. 6.2YTCh. 6 - Prob. 6.3YTCh. 6 - Prob. 6.4YTCh. 6 - Prob. 6.5YTCh. 6 - Prob. 6.6YTCh. 6 - Prob. 6.7YTCh. 6 - Prob. 6.8YTCh. 6 - Prob. 6.9YTCh. 6 - Prob. 6.10YTCh. 6 - Prob. 6.11YTCh. 6 - Prob. 6.12YTCh. 6 - Prob. 6.13YTCh. 6 - Prob. 6.14YTCh. 6 - Prob. 6.15YTCh. 6 - Prob. 6.16YTCh. 6 - Prob. 6.17YTCh. 6 - Prob. 6.18YTCh. 6 - Prob. 6.19YTCh. 6 - Prob. 6.20YT
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- Calculate the pH of a 0.20 M HA solution, given the pKa of HA is 2.22.arrow_forwardNeutralization of phosphate . Given that phosphoricacid (H3PO4) can give up three protons with different pKa values, sketch a plot of pH as a function of added drops of sodium hydroxide solution, starting with a solution ofphosphoric acid at pH 1.0.arrow_forwardDetermine the pH of a 10-3 mol/L solution of (a) hydrogen sulfate ion (pKa = 1.99), (b) formic acid (pKa = 3.75), (c) hypochlorous acid (pKa = 7.53), (d) phenol (pKa = 9.98)arrow_forward
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