40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? (Ka(CH3COOH) = 1.8 × 10−5)
let X = M CH3COOH.
(CH3COOH) = (40mL*XM/60 mL)
(NaOH) added = (20 mL x 0.1M/60mL)
........CH3COOH + NaOH ==> CH3COO^- H2O
I.......(40X/60)...0........0..........0
add...............2/60...............
change...-2/60... -2/60.....2/60.....2/60
equilib (40X-2/60)..0........2/60....2/60
Substitute into the Henderson-Hasselbalch equation and solve for X.
5.10 = 4.74 + log(2/(40x-2) and solve for X
dkls,45
To solve this problem, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base:
pH = pKa + log([A-]/[HA])
Where:
pH = 5.10 (given)
pKa = -log(Ka) = -log(1.8 × 10^(-5)) = 4.74 (given)
[A-] = concentration of the acetate ion (CH3COO-)
[HA] = concentration of the acetic acid (CH3COOH)
We are given that 40.0 mL of acetic acid of unknown concentration is titrated with 0.100 M NaOH. After adding 20.0 mL of the base, the pH is measured.
Step 1: Determine the moles of NaOH added:
The volume of NaOH used is 20.0 mL, and the concentration of NaOH is 0.100 M.
Number of moles of NaOH = volume × concentration = (20.0 mL)(0.100 mol/L) = 0.00200 mol
Step 2: Determine the moles of acetic acid neutralized:
The balanced chemical equation for the reaction between acetic acid and NaOH is:
CH3COOH + NaOH → CH3COONa + H2O
The stoichiometry of the reaction is 1:1, so the moles of acetic acid neutralized are also 0.00200 mol.
Step 3: Calculate the molar concentration of acetic acid:
The volume of the acetic acid solution is 40.0 mL, which is equal to 0.0400 L.
The molar concentration of acetic acid is then:
Molar concentration = moles/volume = 0.00200 mol / 0.0400 L = 0.0500 mol/L
Therefore, the concentration of the original acetic acid solution is 0.0500 mol/L or 0.0500 M.
To find the concentration of the original acetic acid solution, we can use the concept of acid-base titration and the given information.
In this problem, acetic acid (CH3COOH) is a weak acid that can react with sodium hydroxide (NaOH), a strong base. The reaction between these two compounds results in the formation of water and sodium acetate (CH3COONa).
First, let's understand the calculation steps:
1. Determine the number of moles of NaOH used:
- Concentration of NaOH = 0.100 M
- Volume of NaOH used = 20.0 mL = 0.0200 L
- Moles of NaOH used = Concentration x Volume
2. Determine the number of moles of CH3COOH:
- The reaction between NaOH and CH3COOH is 1:1, so the moles of NaOH used are equal to the moles of CH3COOH initially present in the solution.
3. Calculate the initial concentration of CH3COOH:
- Volume of CH3COOH = 40.0 mL = 0.0400 L
- Concentration of CH3COOH = Moles of CH3COOH / Volume of CH3COOH
Now, let's calculate the concentration of the original acetic acid solution step by step:
Step 1: Determine the number of moles of NaOH used:
Moles of NaOH used = 0.100 M x 0.0200 L = 0.00200 mol
Step 2: Determine the number of moles of CH3COOH:
Since the reaction between NaOH and CH3COOH is 1:1, the moles of NaOH used are equal to the moles of CH3COOH initially present in the solution.
Moles of CH3COOH = 0.00200 mol
Step 3: Calculate the initial concentration of CH3COOH:
Concentration of CH3COOH = Moles of CH3COOH / Volume of CH3COOH
Concentration of CH3COOH = 0.00200 mol / 0.0400 L
Concentration of CH3COOH = 0.0500 M
Therefore, the concentration of the original acetic acid solution is 0.0500 M.