Final answer:
The remaining concentration of Ag+ ions after mixing AgNO3 with NaCN and reaching equilibrium will be negligible, as most of the Ag+ will form a stable complex with CN−, which is Ag(CN)2−.
Step-by-step explanation:
The question is asking for the remaining concentration of Ag+ after AgNO3 is mixed with NaCN and reaches equilibrium. The reaction that occurs is between silver ion (Ag+) and cyanide ion (CN−) to form a complex ion, Ag(CN)2−. Given that the initial concentrations are 2.8 x 10−3 M for AgNO3 and 0.10 M for NaCN, we can use stoichiometry to find the amount of Ag+ that reacts with CN− to form Ag(CN)2−. As Ag+ forms a complex with CN−, the concentration of free Ag+ decreases.
To calculate the amount each ion contributes to reaction before reaching equilibrium, we need to take into account the volume and concentration of both solutions. As the concentration of CN− is much higher compared to Ag+ and the Kf for Ag(CN)2− is large, we can assume that all Ag+ reacts with CN−. The remaining concentration of Ag+ would be negligible and close to zero as Ag(CN)2− is very stable and will not dissociate easily to produce free Ag+ ions.