Potassium ferrocyanide is 50% ionised in aqueous solution, its van't Hoff factor will be.......... .

Answer

To find the van't Hoff factor (i) for potassium ferrocyanide (K₄[Fe(CN)₆]) which is 50% ionized in aqueous solution, we can follow these steps:

### Step 1: Understand the dissociation of potassium ferrocyanide
Potassium ferrocyanide dissociates in solution as follows:
\[ K₄[Fe(CN)₆] \rightarrow 4K^+ + [Fe(CN)₆]^{4-} \]
This means that one formula unit of potassium ferrocyanide produces a total of 5 ions (4 potassium ions and 1 ferrocyanide ion).

### Step 2: Define the degree of ionization (α)
The problem states that potassium ferrocyanide is 50% ionized. This means the degree of ionization (α) is 0.5.

### Step 3: Relate the van't Hoff factor (i) to the degree of ionization (α)
The relationship between the van't Hoff factor (i) and the degree of ionization (α) for a substance that dissociates into n ions can be expressed as:
\[ i = 1 + (n - 1)α \]
In this case, n (the total number of ions produced) is 5.

### Step 4: Substitute the values into the equation
Substituting the values into the equation:
\[ i = 1 + (5 - 1) * 0.5 \]
\[ i = 1 + 4 * 0.5 \]
\[ i = 1 + 2 \]
\[ i = 3 \]

### Conclusion
The van't Hoff factor (i) for potassium ferrocyanide when it is 50% ionized in aqueous solution is **3**.

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