`K_(a)` for a monobasic organic acid is `2xx10^(-5)` wiat is pH of 0.2 M aqueous solution of its salt formed with KOH?

Answer

To find the pH of a 0.2 M aqueous solution of the salt formed from a monobasic organic acid and KOH, we can follow these steps:

### Step 1: Understand the Reaction
The salt formed from a monobasic organic acid (let's denote it as HA) and KOH will be of the form K(A). In solution, this salt will dissociate to give A⁻ ions, which can hydrolyze in water.

### Step 2: Write the Hydrolysis Reaction
The hydrolysis of the anion A⁻ can be represented as:
\[ A^- + H_2O \rightleftharpoons HA + OH^- \]

### Step 3: Set Up the Equilibrium Expression
Let \( x \) be the concentration of OH⁻ produced at equilibrium. Initially, the concentration of A⁻ is 0.2 M, and at equilibrium, it will be \( 0.2 - x \). The equilibrium expression for the hydrolysis can be written as:
\[ K_h = \frac{[HA][OH^-]}{[A^-]} \]

### Step 4: Substitute the Values
Since \( [HA] = x \) and \( [OH^-] = x \), we can rewrite the equilibrium expression as:
\[ K_h = \frac{x^2}{0.2 - x} \]

### Step 5: Relate \( K_h \) to \( K_a \)
We know that:
\[ K_h = \frac{K_w}{K_a} \]
Where \( K_w = 1.0 \times 10^{-14} \) and \( K_a = 2 \times 10^{-5} \).

Calculating \( K_h \):
\[ K_h = \frac{1.0 \times 10^{-14}}{2 \times 10^{-5}} = 5.0 \times 10^{-10} \]

### Step 6: Substitute \( K_h \) into the Equilibrium Expression
Now we can set up the equation:
\[ 5.0 \times 10^{-10} = \frac{x^2}{0.2 - x} \]

### Step 7: Make an Approximation
Assuming \( x \) is small compared to 0.2, we can approximate:
\[ 5.0 \times 10^{-10} = \frac{x^2}{0.2} \]
Thus,
\[ x^2 = 5.0 \times 10^{-10} \times 0.2 \]
\[ x^2 = 1.0 \times 10^{-10} \]
\[ x = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \]

### Step 8: Find the Concentration of \( H^+ \)
Using the relation:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9} \]

### Step 9: Calculate the pH
Finally, we can calculate the pH:
\[ pH = -\log[H^+] = -\log(1.0 \times 10^{-9}) = 9 \]

### Final Answer
The pH of the 0.2 M aqueous solution of the salt formed with KOH is **9**.

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