the acid dissociation constant for acetic acid is 1.85x10^-5. Calculate the pH at the equivalent point and the pH at one-half the equivalence point for the titration of 25.00 mL of 0.100 M acetic acid with 0.100 M NaOH?
To calculate the pH at the equivalent point and the pH at one-half the equivalence point for the titration of acetic acid with NaOH, we need to consider the reaction that occurs during titration.
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
At the equivalent point, the moles of acetic acid reacted will be equal to the moles of sodium hydroxide added. This means that all the acetic acid has been neutralized, and we have a solution of sodium acetate (CH3COONa), which is a conjugate base of acetic acid.
At one-half the equivalence point, the moles of acetic acid reacted will be half of the moles of sodium hydroxide added. This means that half of the acetic acid has been neutralized, and we have a solution containing both acetic acid and its conjugate base.
Let's calculate the pH at the equivalent point first:
Step 1: Calculate the moles of acetic acid initially present:
Moles of acetic acid = volume (in L) x concentration
Moles of acetic acid = 0.025 L x 0.100 M = 0.0025 moles
Step 2: Calculate the moles of sodium hydroxide added at the equivalent point:
The balanced equation shows that the ratio between acetic acid and sodium hydroxide is 1:1. So, moles of sodium hydroxide = moles of acetic acid = 0.0025 moles
Step 3: Calculate the concentration of the sodium acetate formed at the equivalent point:
Concentration of sodium acetate = moles of sodium acetate formed / volume (in L)
The volume at equivalent point = volume of acetic acid + volume of sodium hydroxide added
= 0.025 L + 0.025 L = 0.050 L
Concentration of sodium acetate = 0.0025 moles / 0.050 L = 0.050 M
Step 4: Calculate the pOH at the equivalent point:
pOH = -log10 (concentration of sodium acetate)
pOH = -log10 (0.050) = 1.30
Step 5: Calculate the pH at the equivalent point:
pH + pOH = 14
pH = 14 - pOH = 14 - 1.30 = 12.70
Now, let's calculate the pH at one-half the equivalence point:
Step 1: Calculate the moles of sodium hydroxide added at one-half the equivalence point:
The ratio between acetic acid and sodium hydroxide is 1:1, so moles of sodium hydroxide = 0.5 x moles of acetic acid = 0.5 x 0.0025 moles = 0.00125 moles
Step 2: Calculate the moles of acetic acid remaining:
Moles of acetic acid remaining = initial moles of acetic acid - moles of acetic acid reacted
Moles of acetic acid remaining = 0.0025 moles - 0.00125 moles = 0.00125 moles
Step 3: Calculate the concentration of acetic acid and sodium acetate at one-half the equivalence point:
The volume at one-half the equivalence point = volume of acetic acid + volume of sodium hydroxide added
= 0.025 L + 0.0125 L = 0.0375 L
Concentration of acetic acid = moles of acetic acid remaining / volume (in L)
Concentration of acetic acid = 0.00125 moles / 0.0375 L = 0.0333 M
Concentration of sodium acetate = moles of sodium acetate formed / volume (in L)
Concentration of sodium acetate = 0.00125 moles / 0.0375 L = 0.0333 M
Step 4: Calculate the pH at one-half the equivalence point:
pH = -log10 (concentration of acetic acid)
pH = -log10 (0.0333) = 1.48
Therefore, the pH at the equivalent point is 12.70, and the pH at one-half the equivalence point is 1.48.
To calculate the pH at the equivalent point and the pH at one-half the equivalence point for the given titration, you'll need to determine the number of moles of acetic acid and NaOH used at each point.
1. Determine moles of acetic acid:
Moles = concentration (mol/L) × volume (L)
Moles of acetic acid = 0.100 M × 0.02500 L (volume used) = 0.00250 mol
2. Determine moles of NaOH required to reach equivalence point:
Since acetic acid and NaOH react in a 1:1 ratio according to the balanced chemical equation (CH3COOH + NaOH -> CH3COONa + H2O), the moles of NaOH used will be the same as the moles of acetic acid.
Moles of NaOH = 0.00250 mol
3. Calculate the volume of NaOH needed for the equivalence point:
The volume of NaOH required to reach the equivalence point can be determined by dividing the moles of NaOH by its concentration.
Volume of NaOH = Moles of NaOH / Concentration of NaOH
Volume of NaOH = 0.00250 mol / 0.100 M = 0.025 L = 25.0 mL
4. Calculate the pH at the equivalent point:
At the equivalence point, the moles of acetic acid and NaOH react completely, resulting in the formation of sodium acetate (CH3COONa), and no excess acetic acid or NaOH remains. The resulting solution will be a sodium acetate solution.
Since the reaction of acetic acid with NaOH produces equal concentrations of the conjugate base (CH3COO-) and water (H2O), the equivalence point solution will be neutral.
Thus, the pH at the equivalence point will be 7.
5. Calculate the moles of acetic acid used at one-half the equivalence point:
At the one-half equivalence point, half of the moles of acetic acid have reacted. Therefore, the moles of acetic acid remaining will be half of the total initial moles used.
Moles of acetic acid at half equivalence point = 0.00250 mol / 2 = 0.00125 mol
6. Calculate the volume of NaOH needed for the one-half equivalence point:
Using the moles of acetic acid at the one-half equivalence point (0.00125 mol), divide by the concentration of NaOH to find the volume of NaOH needed.
Volume of NaOH = Moles of NaOH / Concentration of NaOH
Volume of NaOH = 0.00125 mol / 0.100 M = 0.0125 L = 12.5 mL
7. Calculate the moles of NaOH remaining at the one-half equivalence point:
Since the reaction between acetic acid and NaOH occurs in a 1:1 ratio, the moles of NaOH at the one-half equivalence point will also be half of the total initial moles used.
Moles of NaOH remaining at half equivalence point = 0.00125 mol
8. Calculate the moles of CH3COONa formed at the one-half equivalence point:
Since acetic acid and NaOH react in a 1:1 ratio, the moles of sodium acetate (CH3COONa) formed will also be equal to the moles of acetic acid used at the one-half equivalence point.
Moles of CH3COONa formed at half equivalence point = 0.00125 mol
9. Calculate the concentration of sodium acetate at the one-half equivalence point:
The volume of the solution at the one-half equivalence point is the sum of the initial volume of acetic acid (25.00 mL) and the volume of NaOH added (12.5 mL).
Total volume = 25.00 mL + 12.5 mL = 37.5 mL = 0.0375 L
Concentration of sodium acetate = Moles of CH3COONa formed / Total volume
Concentration of sodium acetate = 0.00125 mol / 0.0375 L = 0.0333 M
10. Calculate the pH at the one-half equivalence point:
At the one-half equivalence point, the solution will be a mixture of acetic acid and sodium acetate. To calculate the pH, you'll need to consider the acid dissociation constant (Ka) of acetic acid and the concentration of the acetate ion (CH3COO-).
The Ka value for acetic acid is given as 1.85 × 10^-5, which means the acid dissociation reaction can be represented as follows:
CH3COOH ⇌ CH3COO- + H+
To calculate the pH, you'll need to determine the concentration of H+ ions using the equilibrium expression for acetic acid and the concentration of the acetate ion.
[H+] = √(Ka × [CH3COOH] / [CH3COO-])
[H+] = √(1.85 × 10^-5 × 0.00125 M / 0.0333 M)
Then, you can calculate the pH using the equation:
pH = -log10[H+]
Plug in the calculated [H+] value to get the pH at the one-half equivalence point.