If `pK_(a)=-"log"K_(a)=4`, and `K_(a)=Cx^(2-)` then Van't Hoff factor for weak monobasic acid when `C=0.01M` is:

Explanation

To solve the problem, we need to find the Van't Hoff factor (i) for a weak monobasic acid given the dissociation constant (Ka) and concentration (C). Let's break it down step by step.

### Step 1: Understand the dissociation of the weak acid
Let the weak monobasic acid be represented as HA. The dissociation of HA can be represented as:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

### Step 2: Set up the equilibrium expression
Let the initial concentration of HA be \( C \) (which is given as 0.01 M). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- [HA] = \( C(1 - \alpha) \)
- [H⁺] = \( C\alpha \)
- [A⁻] = \( C\alpha \)

### Step 3: Write the expression for Ka
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 4: Relate Ka to the given pKa
We are given that \( pK_a = 4 \). Therefore, we can find \( K_a \):
\[ K_a = 10^{-pK_a} = 10^{-4} \]

### Step 5: Set up the equation
Now we equate the two expressions for \( K_a \):
\[ \frac{C\alpha^2}{1 - \alpha} = 10^{-4} \]
Given \( C = 0.01 \, \text{M} = 10^{-2} \):
\[ \frac{10^{-2}\alpha^2}{1 - \alpha} = 10^{-4} \]

### Step 6: Solve for α
Cross-multiplying gives:
\[ 10^{-2}\alpha^2 = 10^{-4}(1 - \alpha) \]
\[ 10^{-2}\alpha^2 = 10^{-4} - 10^{-4}\alpha \]
Rearranging:
\[ 10^{-2}\alpha^2 + 10^{-4}\alpha - 10^{-4} = 0 \]

### Step 7: Multiply through by \( 10^4 \) to simplify
\[ 10^2\alpha^2 + \alpha - 1 = 0 \]
This is a quadratic equation in \( \alpha \).

### Step 8: Use the quadratic formula
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 100, b = 1, c = -1 \):
\[ \alpha = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 100 \cdot (-1)}}{2 \cdot 100} \]
\[ \alpha = \frac{-1 \pm \sqrt{1 + 400}}{200} \]
\[ \alpha = \frac{-1 \pm 20.024}{200} \]

Calculating the positive root:
\[ \alpha = \frac{19.024}{200} \approx 0.09512 \]

### Step 9: Calculate the Van't Hoff factor (i)
The Van't Hoff factor \( i \) is given by:
\[ i = \frac{\text{Total moles at final state}}{\text{Total moles at initial state}} \]
\[ i = \frac{C(1 - \alpha) + C\alpha + C\alpha}{C} = 1 + \alpha \]
Substituting \( \alpha \):
\[ i = 1 + 0.09512 \approx 1.09512 \]

### Final Answer
The Van't Hoff factor \( i \) for the weak monobasic acid when \( C = 0.01M \) is approximately **1.095**.

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