What is the pH at equivalence point in the titration of 0.1 M `CH_(3)COOH` and 0.1 NaOH? (`K_(a)` for acetic acid is `1.8 xx 10^(-5)`)

Options

A. 7
B. Between 7 and 8
C. Between 8 and 9
D. Between 6 and 7

Answer

To find the pH at the equivalence point in the titration of 0.1 M acetic acid (CH₃COOH) and 0.1 M sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Determine the reaction at the equivalence point
At the equivalence point, all the acetic acid reacts with sodium hydroxide to form sodium acetate (CH₃COONa) and water. The balanced equation is:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Identify the properties of the resulting solution
At the equivalence point, we have a solution of sodium acetate, which is the salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). Sodium acetate will dissociate in water to give acetate ions (CH₃COO⁻).

### Step 3: Calculate the concentration of acetate ions
Since both the acetic acid and sodium hydroxide are of equal concentration (0.1 M) and they react in a 1:1 ratio, the concentration of acetate ions at the equivalence point will be:
\[ \text{Concentration of CH}_3\text{COO}^- = \frac{0.1 \, \text{mol/L} \times V}{2V} = 0.05 \, \text{mol/L} \]
(where V is the volume of the solution at the equivalence point, which is doubled since both reactants are mixed).

### Step 4: Calculate the pKa of acetic acid
Given that the dissociation constant \( K_a \) for acetic acid is \( 1.8 \times 10^{-5} \), we can calculate \( pK_a \):
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 5: Calculate the pKb of acetate ion
Using the relationship \( pK_w = pK_a + pK_b \) (where \( pK_w = 14 \)):
\[ pK_b = pK_w - pK_a = 14 - 4.74 = 9.26 \]

### Step 6: Use the hydrolysis equation to find the pH
The acetate ion (CH₃COO⁻) will undergo hydrolysis in water:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

The equilibrium expression for this reaction can be written as:
\[ K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} \]

Using the concentration of acetate ions (0.05 M) and substituting into the expression:
\[ K_b = \frac{x^2}{0.05 - x} \approx \frac{x^2}{0.05} \]
(where \( x \) is the concentration of \( OH^- \)).

### Step 7: Solve for x (the concentration of OH⁻)
Using \( K_b = 10^{-9.26} \):
\[ 10^{-9.26} = \frac{x^2}{0.05} \]
\[ x^2 = 0.05 \times 10^{-9.26} \]
\[ x^2 = 0.05 \times 5.5 \times 10^{-10} \approx 2.75 \times 10^{-11} \]
\[ x \approx 5.24 \times 10^{-6} \, \text{M} \]

### Step 8: Calculate the pOH and then the pH
Now, calculate the pOH:
\[ pOH = -\log(x) \approx -\log(5.24 \times 10^{-6}) \approx 5.28 \]

Finally, calculate the pH:
\[ pH = 14 - pOH = 14 - 5.28 \approx 8.72 \]

### Final Answer
The pH at the equivalence point in the titration of 0.1 M acetic acid and 0.1 M sodium hydroxide is approximately **8.72**.