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The pH of a solution prepared by dissolving 0.350 mol of acetic acid (CH3CO2H) in 1.00...

The pH of a solution prepared by dissolving 0.350 mol of acetic acid (CH3CO2H) in 1.00 L of water is 2.64. Determine the pH of the solution after adding 0.079 moles of sodium hydroxide (NaOH). The Ka for acetic acid is 1.75E-5 (Assume the final volume is 1.00L)

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Expert Solution

pKa = -log Ka

pKa = -log (1.7 x 10^-5)

pKa = 4.77

CH3COOH   + NaOH ---------------------> CH3COONa + H2O

0.350               0.079                                         0               0 ----------------> initial

0.271               0.0                                            0.079       0.079 --------------> after reaction

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.77 + log (0.079 / 0.271)

pH = 4.24

queen_honey_blossom answered 2 years ago
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