Question
Question: When KI(excess) is added to: (i) \[CuS{{O}_{4}}\](ii) \[HgC{{l}_{2}}\](iii) \[Pb{{(N{{O}_{3}})}_{2...
When KI(excess) is added to:
(i) CuSO4(ii) HgCl2(iii) Pb(NO3)2
(A) a white precipitate of CuI in(i), an orange ppt of HgI2in (ii), and a yellow ppt of PbI2 in (iii) are formed.
(B) a white ppt of Cu2I2 in (i), a red ppt dissolving to HgI42− in (ii), and a yellow ppt. Of PbI2in (iii) are formed.
(C) a white ppt of CuI, HgI2 and PbI2 are formed in each case.
(D) none is correct.
Solution
When KI is added in excess to certain compounds, iodide compounds are formed. Each iodide gives a characteristic precipitate by which we can identify the iodide formed.
Complete step by step solution:
KI reacts with copper sulphate and forms a white precipitate of Cu2I2.
4KI+2CuSO4→Cu2I2↓+I2+2K2SO4
HgCl2 Reacts with KI and forms an orange precipitate of HgI2 which dissolves in excess of KI to give Nessler’s solution.
HgCl2+2KCl→HgI2↓+2KCl
Excess of KI the reaction is HgI2+KI→K2HgI4
When KI reacts with Pb(NO3)2, yellow precipitate is formed of PbI2
2KI+Pb(NO3)2 toPbI2+2KNO3
Thus, when KI is added to CuSO4,HgCl2, and Pb(NO3)2 ,we get a white precipitate of Cu2I2 , an orange ppt of HgI2, and a yellow ppt of PbI2 are formed respectively.
The answer to the question is option (B).
Additional information:
KI or potassium iodide has many applications. It is used in quantitative analysis. It can also be used as a medication as well as a dietary supplement. Potassium iodide is a good reducing agent. The iodide ion is easily oxidised to I2. KI is used in redox titrations. Aqueous KI3 is used as a disinfectant. It is a good fluorescence quenching agent in biomedical research. Potassium iodide has its most important application in organic synthesis, mainly in the preparation of aryl iodides and the reaction is referred to as Sandmeyer’s reaction.
Note: We can get confused with the reaction of KI withHgCl2. In this reaction only in excess of KI Nessler’s solution is formed. Otherwise, an orange precipitate is formed.