The formation constant is $x \times 10^{5}$ for the reaction of a tripositive metal ion with thiocyanate ion to form a mono complex $M^{3 +} + S C N^{-} ⇌ M S C N^{2 +}$ .
If the total metal concentration in the solution is $2.0 \times 10^{- 3}$ M, the total $S C N^{-}$ concentration is $1.50 \times 10^{- 3} M$ , and the free $S C N^{-}$ concentration is $1.0 \times 10^{- 5} M$ . Find the value of x (to nearest integer)

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answer is 3.

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Detailed Solution

$M^{3 +} + S C N^{-} ⇌ M S C N^{2 +}$
$[M S C N^{2 +}] = (1.5 \times 10^{- 3} M) - (1.0 \times 10^{- 5} M) = 1.49 \times 10^{- 3} M$
So, free $[M^{3 +}] = (2.0 \times 10^{- 3} M) - (1.49 \times 10^{- 3} M)$
$= 0.51 \times 10^{- 3} M$
$∴ K = \frac{[M S C N^{2 +}]}{[M^{3 +}] [S C N^{-}]} = \frac{1.49 \times 10^{- 3}}{(0.51 \times 10^{- 3}) \times (1.0 \times 10^{- 5})} = 2.92 \times 10^{- 5} \approx 3.0 \times 10^{5}$