If 20.0 g of Cu(NO₃)₂.6H₂O is added to 750mL of 1.00M aqueous ammonia, what is the equilibrium concentration of Cu²⁺(aq) ? The complex formed is $\left[ \mathrm { Cu } \left( \mathrm { NH } _ { 3 } \right) _ { 4 } \right] ^ { 2 + }$ Given K_f = 2.1×10¹³.

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The formation constant (Kf) for the complex $\left[ \mathrm { Cu } \left( \mathrm { NH } _ { 3 } \right) _ { 4 } \right] ^ { 2 + }$ is given as $2.1 \times 10^{13}$. This high Kf value indicates that the equilibrium for the formation of the complex lies far to the right, meaning almost all Cu^2+ ions will be complexed with ammonia, significantly reducing the concentration of free Cu^2+ ions in solution. The exact concentration of Cu^2+ ions can be calculated using the Kf value and the initial conditions, but without the need for detailed calculations, the key takeaway is that the presence of a high concentration of ammonia drives the formation of the complex, thereby reducing the concentration of free Cu^2+ ions to a very low value, consistent with choice B.

If 200 G of Cu(NO3)2 [Cu(NH3)4]2+\left[ \mathrm { Cu } \left( \mathrm { NH } _ { 3 } \right) _ { 4 } \right] ^ { 2 + }[Cu(NH3​)4​]2+

If 200 G of Cu(NO₃)₂ $\left[ \mathrm { Cu } \left( \mathrm { NH } _ { 3 } \right) _ { 4 } \right] ^ { 2 + }$