The formation constant (K_f) of the complex ‛[Fe(CN)6]4-” is 1.6× $1 0^{17}$ . If the concentration of Fe3+ is 1.0× $1 0^{- 6}$ M and the concentration of CN- is 1.0× $1 0^{- 3}$ M, calculate the concentration of ‛[Fe(CN)6]4-” formed.

medium 2 marks Class 11 · Biology · Chemical Coordination and Integration

Answer

1.6 ×

1 0^{- 7}

The correct answer is 1.6 × $1 0^{- 7}$ M — Result: 1.6× $1 0^{- 7}$ M.

Step-by-step solution

Write the expression for K_f: K_f = [Fe(CN)_ $6^{4 -}$ ]/([ $F e^{3 +}$ ][CN^-]^6).
Rearrange for the concentration of the complex: [Fe(CN)_ $6^{4 -}$ ] = K_f × [ $F e^{3 +}$ ] × [CN^-]^6.
Calculate [CN^-]^6: (1.0× $1 0^{- 3}$ )^6 = 1.0× $1 0^{- 18}$ .
Multiply by [ $F e^{3 +}$ ]: 1.0× $1 0^{- 6}$ × 1.0× $1 0^{- 18}$ = 1.0× $1 0^{- 24}$ .
Multiply by K_f: 1.6× $1 0^{17}$ × 1.0× $1 0^{- 24}$ = 1.6× $1 0^{- 7}$ .
Result: 1.6× $1 0^{- 7}$ M.

Common mistake: Ignoring the sixth power of CN^-

Concept tested: Stability constant calculation