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The neutralization reaction of the weak acid CH₃COOH by strong base NaOH:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
From equation we see 1 mole of acetic acid needs 1 mole of NaOH for neutralization.
The number of moles in 500 ml of 0.167 M NaOH is calculated as:
number of moles = M * V(in liters) = 0.167 x 0.5 L = 0.0835 mole
also for acetic acid:
number of moles = M * V(in liters) = 0.1 x 0.5 L = 0.05 mole
The ICE for neutralization is:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
initial: 0.050 0.0835 0
Change: -0.050 -0.050 0.050
Final: 0 0.0335 0.050
Since the final volume is 1 L, the conc. of each species is equal to its number of moles. so concentration of acetate [CH₃COO⁻] = 0.050 M
Na⁺ ion is spectator ion (it is concentration not change during reaction) so its concentration equal to initial concentration of NaOH = 0.0835 M
The conc. of OH⁻ equal to conc. of remaining NaOH = 0.0335 M
From the relation Kw = [H⁺][OH⁻] we can get [H⁺]
[H⁺] =![\frac{ K_{w}}{[OH^{-} ]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20K_%7Bw%7D%7D%7B%5BOH%5E%7B-%7D%20%5D%7D%20)
= 3.0 x 10⁻¹³ M (note that Kw = 1.0 x 10⁻¹⁴)
The acetate ion formed undergoes hydrolysis as:
CH₃COO⁻(aq) + H₂O(l) → CH₃COOH(aq) + OH⁻(aq)
The ionization constant Kb for this reaction can be written as:
Kb =![\frac{[CH3COOH][OH-]}{[CH3COO-]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH3COOH%5D%5BOH-%5D%7D%7B%5BCH3COO-%5D%7D%20)
Kb = 5.6 x 10⁻¹⁰ and acetate = 0.05M and OH = 0.0335 M
[CH₃COOH] = 8.4 X 10⁻¹⁰ M
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
From equation we see 1 mole of acetic acid needs 1 mole of NaOH for neutralization.
The number of moles in 500 ml of 0.167 M NaOH is calculated as:
number of moles = M * V(in liters) = 0.167 x 0.5 L = 0.0835 mole
also for acetic acid:
number of moles = M * V(in liters) = 0.1 x 0.5 L = 0.05 mole
The ICE for neutralization is:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
initial: 0.050 0.0835 0
Change: -0.050 -0.050 0.050
Final: 0 0.0335 0.050
Since the final volume is 1 L, the conc. of each species is equal to its number of moles. so concentration of acetate [CH₃COO⁻] = 0.050 M
Na⁺ ion is spectator ion (it is concentration not change during reaction) so its concentration equal to initial concentration of NaOH = 0.0835 M
The conc. of OH⁻ equal to conc. of remaining NaOH = 0.0335 M
From the relation Kw = [H⁺][OH⁻] we can get [H⁺]
[H⁺] =
= 3.0 x 10⁻¹³ M (note that Kw = 1.0 x 10⁻¹⁴)
The acetate ion formed undergoes hydrolysis as:
CH₃COO⁻(aq) + H₂O(l) → CH₃COOH(aq) + OH⁻(aq)
The ionization constant Kb for this reaction can be written as:
Kb =
Kb = 5.6 x 10⁻¹⁰ and acetate = 0.05M and OH = 0.0335 M
[CH₃COOH] = 8.4 X 10⁻¹⁰ M