Chemistry
Given that for ethanoic acid pKa=4.75, calculate the pH of sodium ethanoate 0.1M.
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The Ka of ethanoic acid is 1.75x10-5. The Dissociation Equation for ethanoic acid is: CH3CO2H + H2O CH3CO2- + H3O+ Ionization of CH3CO2- + H2O CH3CO2- + H2O H2CO3 + OH- Ka= [H3O+][OH-]/[H2CO3] = 1.75x10-5 The concentration of [H2CO3] can be determined by using the Henderson-Hasselbalch equation. pH = pKa + log([CH3CO2-]/[CH3CO2H]) pH= 4.75 + log(([0.1M]/[0.1M])) pH = 4.75 The pH of 0.1 M Sodium ethanoate is 4.75.
