Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions t — Ionic Equilibrium Chemistry Question
Question
Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is $2 \times 10^{-3}\text{ M}$, the total $\text{SCN}^-$ concentration is $1.51 \times 10^{-3}\text{ M}$ and the free $\text{SCN}^-$ concentration is $1.0 \times 10^{-5}\text{ M}$.
💡 Solution & Explanation
The complexation reaction is $\text{M}^{3+} + \text{SCN}^- \rightleftharpoons \text{M(SCN)}^{2+}$. Given total $\text{SCN}^- = 1.51 \times 10^{-3}\text{ M}$ and free $[\text{SCN}^-] = 1.0 \times 10^{-5}\text{ M}$, the concentration of the complex $[\text{M(SCN)}^{2+}] = (1.51 \times 10^{-3}) - (0.01 \times 10^{-3}) = 1.50 \times 10^{-3}\text{ M}$. The total metal concentration is $2 \times 10^{-3}\text{ M}$, so the free metal concentration $[\text{M}^{3+}] = (2.0 \times 10^{-3}) - (1.50 \times 10^{-3}) = 0.50 \times 10^{-3}\text{ M}$. The formation constant is $K_f = \frac{[\text{M(SCN)}^{2+}]}{[\text{M}^{3+}][\text{SCN}^-]} = \frac{1.50 \times 10^{-3}}{(0.50 \times 10^{-3})(1.0 \times 10^{-5})} = \frac{3}{10^{-5}} = 3 \times 10^5$. Therefore, correct answer is B.