How many millitres of `0.05M K_(4)[Fe(CN)_(6)]` solution is required for titration of `60ml` of `0.01m ZnSO_(4)` solutioin, when the product of reactgion is `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)`?

Explanation

To solve the problem of how many milliliters of `0.05M K4[Fe(CN)6]` solution is required for titration of `60ml` of `0.01M ZnSO4` solution, we will follow these steps:

### Step 1: Determine the normality of the solutions
Normality (N) is related to molarity (M) by the formula:
\[ N = n \times M \]
where \( n \) is the number of equivalents.

For \( K4[Fe(CN)6] \):
- The \( n \) factor is 4 because it can donate 4 moles of \( K^+ \) ions.
- Therefore, the normality of \( K4[Fe(CN)6] \) is:
\[ N_1 = n_1 \times M_1 = 4 \times 0.05 = 0.20 \, N \]

For \( ZnSO4 \):
- The \( n \) factor is 2 because it can donate 2 moles of \( Zn^{2+} \) ions.
- Therefore, the normality of \( ZnSO4 \) is:
\[ N_2 = n_2 \times M_2 = 2 \times 0.01 = 0.02 \, N \]

### Step 2: Use the equivalence formula
Using the formula for titration:
\[ N_1 V_1 = N_2 V_2 \]
where:
- \( N_1 \) is the normality of \( K4[Fe(CN)6] \)
- \( V_1 \) is the volume of \( K4[Fe(CN)6] \) (which we need to find)
- \( N_2 \) is the normality of \( ZnSO4 \)
- \( V_2 \) is the volume of \( ZnSO4 \) (given as 60 mL)

Substituting the known values:
\[ 0.20 \, N \times V_1 = 0.02 \, N \times 60 \, mL \]

### Step 3: Solve for \( V_1 \)
Rearranging the equation to solve for \( V_1 \):
\[ V_1 = \frac{0.02 \times 60}{0.20} \]

Calculating:
\[ V_1 = \frac{1.2}{0.20} = 6 \, mL \]

### Conclusion
Thus, the volume of `0.05M K4[Fe(CN)6]` solution required for the titration is **6 mL**.

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