Ionic EquilibriumHard

Question

Calculate the formation constant for the reaction of a tri-positive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is 2 × 10–3 M, the total SCN– concentration is 1.51 × 10–3 M and the free SCN– concentration is 1.0 × 10–5 M.

Options

A.7.55 × 104
B.3 × 105
C.3.33 × 10−6
D.1.5 × 105

Solution

$M^{3 +} + SCN^{-} \rightleftharpoons M(SCN)^{2 +}$

$2 \times 10^{- 3}M\text{1.51} \times \text{1}\text{0}^{- 3}M0$

Eqn. $2 \times 10^{- 3} - xM\text{1.51} \times \text{1}\text{0}^{- 3} - xMx\text{ M}$

$= 5 \times 10^{- 4}M\text{=1.0} \times \text{1}\text{0}^{- 5}M\text{=1.5} \times \text{1}\text{0}^{- 3}\text{ M}$

$\therefore K_{f} = \frac{1.5 \times 10^{- 3}}{5 \times 10^{- 4} \times 1 \times 10^{- 5}} = 3 \times 10^{5}$

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