If enthalpy of neutralization of CH3COOH by NaOH is –49.86 kJ/mol then enthalpy of ionzation of CH3COOH is:
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ΔH = ΔH° – ΔH° (H + OH– → H2O )
= – 49.86 – (–55.84) kJ/mole
= 5.98 kJ/mole
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