Bonding between metal and ligand

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Question

How many milliliters of 0.05M $K_{4} [F e {(C N)}_{6}]$ solution are required for titration of 60 ml of 0.01 M ${ZnSO}_{4}$ solution, when the product of reaction is $K_{2} Z n_{3} {[F e {(C N)}_{6}]}_{2} ?$

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Solution

$4 k_{4} [F e {(C N)}_{6}] + 6 Z n S o_{4} \to 2 K_{2} Z n_{3} {[F e {(C N)}_{6}]}_{2} + 6 K_{2} S o_{4}$
$m i |m a| k_{4} [F e {(C N)}_{6}] = m i |m a| z n S o_{4}$
$4 \to 6$
$x \times 0.05 \to 60 \times 0.1$
$x = \frac{4 \times 60 \times 0.1}{0.05 \times 6} = 8$

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