Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
