The ionization constant of chloroacetic acid is `1.35xx10^(-3)`. What will be the `pH` of `0.1 M` acid and its `0.1M` sodium salt solution?

Explanation

To solve the problem of finding the pH of a 0.1 M solution of chloroacetic acid and its 0.1 M sodium salt solution, we can follow these steps:

### Step 1: Determine the ionization of chloroacetic acid
Chloroacetic acid (HA) ionizes in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
The ionization constant (Ka) is given as \( 1.35 \times 10^{-3} \).

### Step 2: Set up the expression for Ka
For a weak acid, the expression for the ionization constant is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Assuming that the initial concentration of the acid (C) is 0.1 M and letting \( \alpha \) be the degree of ionization, we can express the concentrations at equilibrium:
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)
- \([HA] = C(1 - \alpha)\)

Substituting these into the Ka expression gives:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Step 3: Solve for the concentration of H+
Assuming \( \alpha \) is small (which is valid for weak acids), we can neglect \( \alpha \) in the denominator:
\[ K_a \approx \frac{C\alpha^2}{1} \]
Thus:
\[ \alpha^2 = \frac{K_a}{C} \]
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

Substituting the values:
\[ \alpha = \sqrt{\frac{1.35 \times 10^{-3}}{0.1}} = \sqrt{0.0135} \approx 0.116 \]

### Step 4: Calculate the concentration of H+
Now, we can find the concentration of \( H^+ \):
\[ [H^+] = C\alpha = 0.1 \times 0.116 = 0.0116 \, M \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value we found:
\[ pH = -\log(0.0116) \approx 1.93 \]

### Step 6: Calculate the pH of the sodium salt solution
For the sodium salt of chloroacetic acid (A^-), it will undergo hydrolysis:
\[ A^- + H_2O \rightleftharpoons HA + OH^- \]

### Step 7: Set up the hydrolysis constant (Kh)
The hydrolysis constant \( K_h \) can be calculated using:
\[ K_h = \frac{K_w}{K_a} \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C. Thus:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.35 \times 10^{-3}} \approx 7.41 \times 10^{-12} \]

### Step 8: Calculate the concentration of OH-
Using the hydrolysis constant:
\[ K_h = \frac{[HA][OH^-]}{[A^-]} \]
Assuming \( x \) is the concentration of \( OH^- \) produced, we have:
\[ K_h \approx \frac{x^2}{C} \]
Thus:
\[ x^2 = K_h \cdot C \]
\[ x = \sqrt{K_h \cdot C} = \sqrt{7.41 \times 10^{-12} \cdot 0.1} \approx 8.61 \times 10^{-6} \]

### Step 9: Calculate the pOH
Now, we can find the pOH:
\[ pOH = -\log[OH^-] \]
\[ pOH = -\log(8.61 \times 10^{-6}) \approx 5.06 \]

### Step 10: Calculate the pH
Finally, we can find the pH:
\[ pH = 14 - pOH = 14 - 5.06 = 8.94 \]

### Final Answers
- The pH of the 0.1 M chloroacetic acid solution is approximately **1.93**.
- The pH of the 0.1 M sodium salt solution is approximately **8.94**.