A solution is made by mixing exactly 500 mL of 0.124 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 × 10−5
[H+]
× 10
Enter your answer in scientific notation.
[OH−]
[CH3COOH]
× 10
Enter your answer in scientific notation.
[Na+]
[CH3COO−]
We will assume that the volumes are additive.
mols NaOH initially = M x L = 0.124 x 0.500 = 0.062
mols CH3COOH initially = 0.100 x 0.500 = 0.05
The ICE chart looks like this.
........NaOH + CH3OOH ==> CH3COONa + H2O
I.......0.062........0.05...................0...................0
C.........-0.05.....-0.05.................0.05...........0.05
E.......0.012...........0...................0.05..........0.05
From the E line, (Na^+) = 0.012 from NaOH left + 0.05 from CH3COOH formed = 0.062 mols and that is in 1 L assume the volumes are additive so (Na^+) = mols/L = ?
From the E line, (CH3COO^-) = 0.05 mols/L = ?
From the E line, and ignoring any hydrolysis of the acetate ion,
(CH3COOH) = 0
(H^+) is calculated from
(H^+)(OH^-) = Kw = 1E-14.
The (OH^-) from the E line is 0.012/1L = ? and plug that into the Kw expression and solve for H^+.
Check this thoroughly.
To calculate the equilibrium concentration of the species, we need to use the principles of stoichiometry and the equilibrium constant expression for the reaction between NaOH and CH3COOH.
The balanced chemical equation for the reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
Given:
- Volume of NaOH solution = 500 mL = 0.500 L
- Volume of CH3COOH solution = 500 mL = 0.500 L
- Concentration of NaOH solution = 0.124 M
- Concentration of CH3COOH solution = 0.100 M
- Ka of CH3COOH = 1.8 × 10−5
Step 1: Calculate the moles of NaOH and CH3COOH.
moles of NaOH = concentration * volume = 0.124 M * 0.500 L = 0.0620 moles
moles of CH3COOH = concentration * volume = 0.100 M * 0.500 L = 0.0500 moles
Step 2: Determine the limiting reactant.
Since the reaction is a 1:1 stoichiometric ratio between NaOH and CH3COOH, the limiting reactant is the one with the lesser number of moles. In this case, CH3COOH is the limiting reactant.
Step 3: Calculate the moles of products formed.
moles of CH3COONa = moles of CH3COOH (because the reaction is 1:1)
moles of CH3COONa = 0.0500 moles
Step 4: Calculate the concentrations of the species at equilibrium.
[H+] = [CH3COO-] = moles of CH3COONa / total volume of solution
[H+] = [CH3COO-] = 0.0500 moles / (0.500 L + 0.500 L) = 0.0500 M
[OH-] = Kw / [H+]
Kw (ion product of water) = 1.0 × 10^-14 at 25°C
[OH-] = 1.0 × 10^-14 / 0.0500 M = 2.0 × 10^-13 M
[CH3COOH] = moles of CH3COOH / total volume of solution
[CH3COOH] = 0.0500 moles / (0.500 L + 0.500 L) = 0.0500 M
[Na+] = moles of CH3COONa / total volume of solution
[Na+] = 0.0500 moles / (0.500 L + 0.500 L) = 0.0500 M
Therefore, the equilibrium concentrations of the species are:
[H+] = 5.00 × 10^-2 M
[OH-] = 2.0 × 10^-13 M (in scientific notation)
[CH3COOH] = 5.00 × 10^-2 M
[Na+] = 5.00 × 10^-2 M
[CH3COO-] = 5.00 × 10^-2 M