See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The Henderson-Hasselbalch equation relates pH, pKa, and the ratio of deprotonated to protonated species. When pH > pKa, the deprotonated (conjugate base) form predominates. When pH < pKa, the protonated (acid) form predominates. Step 1: Analyze acetic acid (CH3COOH, pKa = 4.8) at pH 7.0. Using Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) 7.0 = 4.8 + log([CH3CO2-]/[CH3CO2H]) log([CH3CO2-]/[CH3CO2H]) = 2.2 [CH3CO2-]/[CH3CO2H] = 10^2.2 ≈ 158 Since pH (7.0) >> pKa (4.8), the deprotonated form CH3CO2- overwhelmingly predominates. The major species is the acetate ion, CH3CO2-. Step 2: Analyze ethanol (CH3CH2OH, pKa = 16.0) at pH 7.0. Using Henderson-Hasselbalch: 7.0 = 16.0 + log([CH3CH2O-]/[CH3CH2OH]) log([CH3CH2O-]/[CH3CH2OH]) = -9.0 [CH3CH2O-]/[CH3CH2OH] = 10^-9 Since pH (7.0) << pKa (16.0), the protonated form CH3CH2OH overwhelmingly predominates. The major species is ethanol itself, CH3CH2OH. Step 3: Identify the major species. Major species: CH3CO2- (acetate) and CH3CH2OH (ethanol). Why other options fail: (a) CH3CO2H and CH3CO2OH - incorrect; acetic acid would be mostly deprotonated at pH 7, and CH3CO2OH is not a meaningful species here. (b) CH3CH2O- and CH3CH2OH - incorrect; ethoxide (CH3CH2O-) would not be the major form at pH 7 since pKa of ethanol is 16. (c) CH3CO2H and CH3CH2O- - incorrect on both counts; acetic acid is mostly deprotonated at pH 7, and ethoxide is not significant at pH 7. (d) CH3CO2- and CH3CH2OH - correct; acetate predominates (pH > pKa of acetic acid) and ethanol predominates (pH < pKa of ethanol). Therefore, the correct answer is D.