The Ka expression for the dissociation of acetic acid in water is based on the following equilibrium at 25 ° C: HC2H3O2 (l) + H2O H + (aq) + C2H3O2- (aq) What will the ΔG ° be if Ka = 1.8 x 10-5?
Added by Joseph H.
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314 J/mol*K) and T is the temperature in Kelvin (25°C = 298 K). Show more…
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In the reaction CH3COOH + H2O ⇌ H3O+ + CH3COO- the base constant Ka of acetic acid CH3COOH is approximately 10-5 at 25 °C . What is the pKb of the acetate ion, CH3COO-?
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The dissociation of acetic acid, CH3COOH, has an equilibrium constant at 25 °C of 1.8 x 10^-5. The reaction is CH3COOH (aq) ⇌ CH3COO- (aq) + H+(aq). If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of water and that of CH3COO- is 8.1 x 10^-3 moles in the same 0.500 L, calculate [H+] for the reaction.
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A $0.0560-\mathrm{g}$ quantity of acetic acid is dissolved in enough water to make $50.0 \mathrm{mL}$ of solution. Calculate the concentrations of $\mathrm{H}^{+}, \mathrm{CH}_{3} \mathrm{COO}^{-},$ and $\mathrm{CH}_{3} \mathrm{COOH}$ at equilibrium. ($K_{\mathrm{a}}$ for acetic acid $=1.8 \times 10^{-5}$).
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