If Pb^{2+}, Cu^{2+}, Al^{3+}, Cd^{2+}, Zn^{2+ | vedclass

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

Question

(B) $1$. When $H^{+}/H_2S$ is added,$Pb^{2+}$,$Cu^{2+}$,and $Cd^{2+}$ precipitate as their respective sulfides ($PbS$,$CuS$,$CdS$) because they belong

Vedclass · Accepted Answer

Clear solution in which two radicals are present as a soluble complex

Vedclass · Answer

(B) $1$. When $H^{+}/H_2S$ is added,$Pb^{2+}$,$Cu^{2+}$,and $Cd^{2+}$ precipitate as their respective sulfides ($PbS$,$CuS$,$CdS$) because they belong to Group $II$ of the qualitative analysis scheme.$2$. The remaining ions in solution $(X)$ are $Al^{3+}$ and $Zn^{2+}$.$3$. When excess $NaOH$ is added to solution $(X)$ containing $Al^{3+}$ and $Zn^{2+}$,both form soluble complexes.$4$. $Al^{3+} + 4OH^{-} \rightarrow [Al(OH)_4]^{-}$ (soluble aluminate).$5$. $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_4]^{2-}$ (soluble zincate).$6$. Thus,a clear solution is obtained containing two radicals as soluble complexes.

If $Pb^{2+}, Cu^{2+}, Al^{3+}, Cd^{2+}, Zn^{2+}$ radicals are present in an aqueous solution,and it reacts with $H^{+}/H_2S$,a precipitate is produced. The solution $(X)$ obtained after the removal of the precipitate is treated with excess $NaOH$. What is the result?

Similar Questions

$A$ colourless crystalline salt $X$ is soluble in dilute $HCl$. On adding $NaOH$ solution,it gives a white precipitate which is insoluble in excess of $NaOH$. $X$ is

Which of the following gives a black precipitate when $H_2S$ gas is passed through its solution?

On heating $NaCl + K_2Cr_2O_7 + \text{conc. } H_2SO_4$,the gas evolved is:

Which reagent can be used to identify nickel ion $(Ni^{2+})$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module