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

To calculate the equilibrium concentrations of H, HCOOH, CH3COO-, OH-, and Na+, follow these steps: calculate the concentration of CH3COOH, calculate the concentration of HCOO- and Na+, calculate the concentration of OH-, and finally calculate the concentration of H+ using the equation Kw = [H+][OH-].

Step-by-step explanation:

To calculate the equilibrium concentration of H, HCOOH, CH3COO-, OH-, Na+ in the solution, we need to follow these steps:

1. Calculate the concentration of CH3COOH using the number of millimoles of CH3COOH and the total volume of the solution at this point in the titration. Using the given values, the concentration of CH3COOH is 7.27 x 10^-2 M.
2. Calculate the concentration of HCOO- (acetate ion) using the amount of CH3COOH that has reacted with NaOH. Since they react in a 1:1 stoichiometry, the final concentration of CH3COO- is 1.00 mmol.
3. Calculate the concentration of Na+ (sodium ion) using the same method as in step 2. The final concentration of Na+ is 1.00 mmol.
4. Calculate the OH- concentration using the amount of NaOH added. Since NaOH reacts in a 1:1 ratio with CH3COOH, the amount of excess CH3COOH is 4.00 mmol. Therefore, the final concentration of OH- is 1.00 mmol.
5. Finally, calculate the concentration of H+ (H3O+) using the equation Kw = [H+][OH-]. The concentration of H+ is 5.3 x 10^-6 M.

Answer 2

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⁺] =
`( 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 =
`([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