M^{n+}(aq.) + KI to Xdownarrow text{ (ppt.)} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. $Pb^{2+} + 2KI 	o PbI_2\downarrow$ (yellow ppt.). $PbI_2$ is soluble in excess $KI$ due to the formation of a soluble complex $[PbI_4]^{2-}$

Vedclass · Accepted Answer

$Cu^{2+}(aq.)$

Vedclass · Answer

(B) $1$. $Pb^{2+} + 2KI 	o PbI_2\downarrow$ (yellow ppt.). $PbI_2$ is soluble in excess $KI$ due to the formation of a soluble complex $[PbI_4]^{2-}$.$2$. $Cu^{2+} + 2KI 	o CuI_2 	o Cu_2I_2\downarrow + I_2$. $Cu_2I_2$ is insoluble in excess $KI$.$3$. $Bi^{3+} + 3KI 	o BiI_3\downarrow$ (black ppt.). $BiI_3$ is soluble in excess $KI$ due to the formation of $[BiI_4]^-$.$4$. $Hg^{2+} + 2KI 	o HgI_2\downarrow$ (red ppt.). $HgI_2$ is soluble in excess $KI$ due to the formation of $[HgI_4]^{2-}$.Therefore,$Cu^{2+}$ is the cation that forms a precipitate which remains insoluble in excess $KI$.

$M^{n+}(aq.) + KI \to X\downarrow \text{ (ppt.)}$ $\xrightarrow{Excess KI} \text{ppt. remains insoluble in excess } KI \text{ solution. Then cation } M^{n+}(aq.) \text{ can be}$

Similar Questions

Which one of the following statements is correct?

Concentrated aqueous $NaOH$ can separate a mixture of

Which of the following sulfides is yellow in color?

$A$ light greenish coloured salt was soluble in water. On passing $H_2S$ into the solution,a black precipitate was obtained which dissolved readily in $HCl$. The metal ion present is:

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$Fe(OH)_3 \downarrow + NaOH(\text{excess}) \longrightarrow \text{No reaction}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module