Explain, with balanced ionic equations, why
a. CuI2 dissolves in ammonia solution
b. AgBr dissolves in NaCN solution,
c. HgCl2 dissolves in KCL solution.
a. When CuI2 (copper(II) iodide) dissolves in ammonia solution, it undergoes complexation with ammonia to form a soluble complex ion. The balanced ionic equation for this reaction can be written as:
CuI2(s) + 4NH3(aq) → [Cu(NH3)4]I2(aq)
b. When AgBr (silver bromide) dissolves in NaCN (sodium cyanide) solution, it forms a soluble complex ion known as dicyanoargentate(I) ion. The balanced ionic equation for this reaction can be written as:
AgBr(s) + 2NaCN(aq) → Na2[Ag(CN)2](aq) + NaBr(aq)
c. When HgCl2 (mercury(II) chloride) dissolves in KCl (potassium chloride) solution, it forms a soluble complex ion known as tetrachloromercurate(II) ion. The balanced ionic equation for this reaction can be written as:
HgCl2(s) + 2KCl(aq) → 2K[HgCl4](aq)
a. To explain why CuI2 (copper(II) iodide) dissolves in ammonia solution, we need to determine the reaction of CuI2 with ammonia (NH3). Firstly, let's write the dissociation equation for CuI2 in water:
CuI2 (s) → Cu2+ (aq) + 2I- (aq)
Now, when CuI2 is dissolved in ammonia solution, it forms complex ions with ammonia. Ammonia molecules can act as a Lewis base, providing an electron pair to form coordination complexes. The balanced ionic equation for the dissolution of CuI2 in ammonia solution can be written as:
CuI2 (s) + 4NH3 (aq) → [Cu(NH3)4]2+ (aq) + 2I- (aq)
This equation shows that CuI2 dissociates into Cu2+ ions, which complex with four ammonia molecules to form the copper(II) tetraammine complex ion. The iodide ions (I-) remain dissociated.
b. AgBr (silver bromide) dissolves in NaCN (sodium cyanide) solution due to the formation of a soluble silver complex. Firstly, let's write the dissociation equation for AgBr in water:
AgBr (s) → Ag+ (aq) + Br- (aq)
Now, when AgBr is dissolved in NaCN solution, the cyanide ion (CN-) acts as a ligand, forming a stable complex with silver ions. The balanced ionic equation for the dissolution of AgBr in NaCN solution can be written as:
AgBr (s) + 2CN- (aq) → [Ag(CN)2]- (aq) + Br- (aq)
This equation shows that AgBr dissociates into Ag+ ions, which form a soluble complex with two cyanide ions. The bromide ions (Br-) remain dissociated.
c. HgCl2 (mercury(II) chloride) dissolves in KCl (potassium chloride) solution due to the formation of a soluble mercury complex. Firstly, let's write the dissociation equation for HgCl2 in water:
HgCl2 (s) → Hg2+ (aq) + 2Cl- (aq)
Now, when HgCl2 is dissolved in KCl solution, the chloride ions (Cl-) from KCl ions and mercury ions (Hg2+) form a soluble complex. The balanced ionic equation for the dissolution of HgCl2 in KCl solution can be written as:
HgCl2 (s) + 2K+ (aq) + 2Cl- (aq) → [HgCl4]2- (aq) + 2K+ (aq)
This equation shows that HgCl2 dissociates into Hg2+ ions, which form a soluble complex with four chloride ions, resulting in the formation of the tetrachloromercury(II) complex ion. The potassium ions (K+) remain dissociated.
Cu^+2 + 2I^- + 4NH3 ==> Cu(NH3)^+2 + 2I^-
etc.