The degree of dissociation of weak electrolyde is inversely proportional to the square root fo concentration. It is called Ostwald's dilution law.
`alpha = sqrt((K_(a))/(c))` As the tempertaure increases, degree of dissociation will increase.
`(alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2))))` if concentration is same.
`(alpha_(1))/(alpha_(2)) = sqrt((c_(2))/(c_(1)))` if acid is same.
`pH` of `0.005 M HCOOH [K_(a) = 2 xx 10^(-4)]` is equal to

Options

A. `3`
B. `2`
C. `4`
D. `5`

Explanation

To find the pH of a 0.005 M solution of formic acid (HCOOH) with a dissociation constant \( K_a = 2 \times 10^{-4} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of formic acid in water can be represented as:
\[ \text{HCOOH} \rightleftharpoons \text{HCOO}^- + \text{H}_3\text{O}^+ \]

### Step 2: Set up the expression for the dissociation constant \( K_a \)
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{HCOO}^-][\text{H}_3\text{O}^+]}{[\text{HCOOH}]} \]

### Step 3: Define the initial concentrations and changes
Let the degree of dissociation be \( \alpha \). Initially, the concentration of HCOOH is 0.005 M. At equilibrium:
- The concentration of HCOOH will be \( 0.005(1 - \alpha) \)
- The concentration of HCOO⁻ and H₃O⁺ will both be \( 0.005\alpha \)

### Step 4: Substitute into the \( K_a \) expression
Substituting the equilibrium concentrations into the \( K_a \) expression:
\[ K_a = \frac{(0.005\alpha)(0.005\alpha)}{0.005(1 - \alpha)} = \frac{0.005^2 \alpha^2}{0.005(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{0.005\alpha^2}{1 - \alpha} \]

### Step 5: Substitute the value of \( K_a \)
Now substitute \( K_a = 2 \times 10^{-4} \):
\[ 2 \times 10^{-4} = \frac{0.005\alpha^2}{1 - \alpha} \]

### Step 6: Rearranging the equation
Multiplying both sides by \( (1 - \alpha) \):
\[ 2 \times 10^{-4} (1 - \alpha) = 0.005\alpha^2 \]
Expanding this gives:
\[ 2 \times 10^{-4} - 2 \times 10^{-4} \alpha = 0.005\alpha^2 \]
Rearranging leads to:
\[ 0.005\alpha^2 + 2 \times 10^{-4} \alpha - 2 \times 10^{-4} = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 0.005 \), \( b = 2 \times 10^{-4} \), and \( c = -2 \times 10^{-4} \).

Calculating the discriminant:
\[ b^2 - 4ac = (2 \times 10^{-4})^2 - 4(0.005)(-2 \times 10^{-4}) = 4 \times 10^{-8} + 0.00004 = 4.04 \times 10^{-8} \]

Now substituting into the quadratic formula:
\[ \alpha = \frac{-2 \times 10^{-4} \pm \sqrt{4.04 \times 10^{-8}}}{2 \times 0.005} \]
Calculating the square root:
\[ \sqrt{4.04 \times 10^{-8}} \approx 2.01 \times 10^{-4} \]
Thus:
\[ \alpha = \frac{-2 \times 10^{-4} + 2.01 \times 10^{-4}}{0.01} \approx 0.181 \]

### Step 8: Calculate \([H_3O^+]\)
The concentration of \( [H_3O^+] \) is:
\[ [H_3O^+] = 0.005\alpha = 0.005 \times 0.181 \approx 9.05 \times 10^{-4} \text{ M} \]

### Step 9: Calculate the pH
Using the formula for pH:
\[ \text{pH} = -\log[H_3O^+] = -\log(9.05 \times 10^{-4}) \]
Using logarithmic properties:
\[ \text{pH} = -\log(9.05) - \log(10^{-4}) \approx -0.956 + 4 \approx 3.044 \]
Thus, the pH is approximately 3.

### Final Answer
The pH of the 0.005 M HCOOH solution is approximately **3**.