Tittle-Normal Titration Curve of Acetic Acid with Sodium Hydroxide

In this article, we will walk through the detailed calculations and steps required to draw the titration curve for the reaction of acetic acid (CH3COOH) with sodium hydroxide (NaOH). The pKa of acetic acid is given as 4.67. We will determine the initial pH, the buffer zone pH range, the midpoint of the buffer zone pH, the equivalence point pH, and the final pH. Subsequently, we will outline the steps to draw the titration curve and its typical S-shape.

1. Initial pH

The initial solution consists of 20 mL of 0.100 M acetic acid. To find the initial pH, we use the formula for the dissociation of a weak acid:

$$pH pKa - logleft(frac{[A^-]}{[HA]}right)$$

At the start, no base is present, hence:

$$[A^-] 0$$

The concentration of acetic acid, [HA], is 0.100 M. Using the dissociation constant (K_a) of acetic acid:

$$K_a 10^{-pKa} 10^{-4.76} approx 1.74 times 10^{-5}$$

The equilibrium expression is:

$$K_a frac{[H^ ][A^-]}{[HA]}$$

Let (x) be the concentration of (H^ ) produced. At equilibrium:

$$[H^ ] x, [A^-] x, [HA] 0.100 - x approx 0.100$$

Therefore, the equation becomes:

$$1.74 times 10^{-5} frac{x^2}{0.100}$$

Solving for (x):

$$x^2 1.74 times 10^{-6} implies x approx 0.00132 M$$

Now calculate the pH:

$$pH -log(0.00132) approx 2.88$$

2. Buffer Zone pH Range

The buffer zone occurs before the equivalence point where both the weak acid and its conjugate base are present. The pH in the buffer zone can be estimated using the Henderson-Hasselbalch equation:

$$pH pKa - logleft(frac{[A^-]}{[HA]}right)$$

During the titration, as NaOH is added, the concentration of acetate ion ([A^-]) increases while ([HA]) decreases. At the midpoint of the buffer zone, where half of the acetic acid has been neutralized, ([A^-] [HA]), thus:

$$pH pKa 4.67$$

3. Midpoint of Buffer Zone pH

The pH at the midpoint occurs when 10 mL of NaOH is added, neutralizing 10 mL of acetic acid. The concentrations of acetic acid and acetate are equal:

$$pH pKa 4.67$$

4. Equivalence Point

At the equivalence point, all acetic acid has been converted to acetate. The total volume at this point is 40 mL. The concentration of acetate ion at the equivalence point is:

$$[A^-] frac{0.100 M times 20 mL}{40 mL} 0.050 M$$

Since acetate (A^-) is a weak base, we use the formula to calculate the pH using the conjugate base constant (K_b):

$$K_b cdot K_a K_w implies K_b frac{K_w}{K_a}$$

Given (K_w 1.0 times 10^{-14}) and (K_a 1.74 times 10^{-5}), then:

$$K_b approx frac{1.0 times 10^{-14}}{1.74 times 10^{-5}} approx 5.75 times 10^{-10}$$

Using (K_b) to find ([OH^-]):

$$K_b frac{[OH^-]^2}{[A^-]} implies [OH^-]^2 K_b times [A^-] 5.75 times 10^{-10} times 0.050$$

$$[OH^-]^2 approx 2.875 times 10^{-11} implies [OH^-] approx 5.37 times 10^{-6} M$$

Calculating pOH:

$$pOH -log(5.37 times 10^{-6}) approx 5.27$$

Calculating pH:

$$pH 14 - pOH approx 14 - 5.27 approx 8.73$$

5. Final pH

The final pH after adding enough NaOH to reach the equivalence point is approximately:

$$pH approx 8.73$$

Summary of Results

Initial pH: 2.88 Buffer Zone pH Range: 4.67 (midpoint) Midpoint of Buffer Zone pH: 4.67 Equivalence Point pH: 8.73 Final pH: 8.73

Titration Curve

To draw the titration curve:

Start at pH 2.88. Gradually increase pH as NaOH is added. The pH will rise slowly in the buffer zone around pH 4.67. At the equivalence point (20 mL NaOH added), the pH jumps to around 8.73. Beyond the equivalence point, the pH rises more steeply.

You can plot these points on a graph to visualize the titration curve, which will show a characteristic S-shape with a steep rise at the equivalence point.