A solution is made by mixing exactly 500 mL of 0.156 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 M
Enter your answer in scientific notation.
[OH−]
M
[CH3COOH]
× 10 M
Enter your answer in scientific notation.
[Na+]
M
[CH3COO−]
M
Let's call CH3COOH simply HAc. Saves typing that way.
.
millimols HAc initially = mL x M = 500 x 0.1 = 50
millimols NaOH initially = 500 x 0.156 = 78
........HAc + NaOH --> NaAc + H2O
I.........50..........78...........0...........0
C......-50..........-50.........+50.......50
E.........0...........28..........50..........50
Concentration of the species is millimols/mL. Remember mL is now 1000 because 500 mL has been added to 500 mL to make 1000 mL.
(HAc) = 0
(OH^-) = 28/1000 = ?
(H^+)(OH^-) = Kw. You know Kw and OH, solve for H.
(Na^+) is from the NaOH
(Ac^-) is from the NaAc
Etc
Post your work if you get stuck.
To calculate the equilibrium concentration of the species, we need to determine the reaction that occurs when NaOH and CH3COOH are mixed. NaOH is a strong base, and CH3COOH is a weak acid. They will react to form water and the sodium acetate salt (CH3COONa).
The balanced equation for the reaction is as follows:
CH3COOH + NaOH → CH3COONa + H2O
To solve this problem, we can use the concept of stoichiometry and the given values:
1. Calculate the number of moles of NaOH and CH3COOH:
Molarity (M) = moles/volume (L)
For NaOH:
Molarity = 0.156 M
Volume = 500 mL = 0.5 L
Moles of NaOH = (0.156 M)(0.5 L) = 0.078 mol
For CH3COOH:
Molarity = 0.100 M
Volume = 500 mL = 0.5 L
Moles of CH3COOH = (0.100 M)(0.5 L) = 0.05 mol
2. From the balanced equation, we can see that NaOH and CH3COOH react in a 1:1 molar ratio.
Since 0.05 mol of CH3COOH was used, 0.05 mol of NaOH will react.
3. Calculate the moles of each species at equilibrium:
[H+]: The acid, CH3COOH, has donated a proton (H+), so [H+] = 0.05 mol.
[OH-]: The base, NaOH, has consumed a hydroxide ion (OH-), so [OH-] = 0.05 mol.
[CH3COOH]: The amount of CH3COOH remaining at equilibrium is zero since it has reacted completely.
[Na+]: The amount of Na+ formed at equilibrium is 0.05 mol.
[CH3COO-]: The amount of CH3COO- formed at equilibrium is 0.05 mol.
4. Convert the molar concentrations to scientific notation:
[H+]: [H+] = 0.05 mol = 5.0 × 10^-2 M
[OH-]: [OH-] = 0.05 mol = 5.0 × 10^-2 M
[CH3COOH]: [CH3COOH] = 0 mol = 0.0 × 10^-2 M
[Na+]: [Na+] = 0.05 mol = 5.0 × 10^-2 M
[CH3COO-]: [CH3COO-] = 0.05 mol = 5.0 × 10^-2 M
Therefore, the equilibrium concentrations are:
[H+]: 5.0 × 10^-2 M
[OH-]: 5.0 × 10^-2 M
[CH3COOH]: 0.0 × 10^-2 M
[Na+]: 5.0 × 10^-2 M
[CH3COO-]: 5.0 × 10^-2 M