Question

A solution is made mixing 5.0 x 10^2 ml of .167 m naoh with 5.00 x 10^2 ml of 0.100 m ch3cooh. calculate the equilibrium concentration of h , hcooh, ch3coo-, oh-, na _

Answer 1

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⁺] = $\frac{ K_{w}}{[OH^{-} ]}$
= 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 = $\frac{[CH3COOH][OH-]}{[CH3COO-]}$
Kb = 5.6 x 10⁻¹⁰ and acetate = 0.05M and OH = 0.0335 M
[CH₃COOH] = 8.4 X 10⁻¹⁰ M