6-2.Equilibrium-II (Ionic Equilibrium)
medium

એક દ્રાવણો જે  ${10^{ - 3}}M$ દરેકમાં  $M{n^{2 + }},\,F{e^{2 + }},\,Z{n^{2 + }}$ અને $H{g^{2 + }}$ ની પ્રક્રિયા ${10^{ - 16}}M$ સલ્ફાઇડ આયનથી કરવામાં આવે છે.જો $MnS,\,FeS,\,ZnS$ અને $HgS$નો ${K_{sp}}$ના મૂલ્યો અનુક્રમે ${10^{ - 15}},\,{10^{ - 23}},\,{10^{ - 20}}$ અને ${10^{ - 54}}$, જે પ્રથમ અવક્ષેપન થશે?

A

$FeS$

B

$MgS$

C

$HgS$

D

$ZnS$

(IIT-2003)

Solution

Ionic product in the solution $=10^{-3} \times 10^{-16}=10^{-19}$. The metal sulphide having the lowest solubility will precipitate first provided the ionic product is higher than the $K_{\text {sp. }}$. Here, all salts are of the same valence type. So, the sulphide having the lowest $K_{s p}$ value will precipitate first provided $K_{s p}\,<\,10^{-19}$. HgS has the lowest $K _{\text {sp }}$ value $\left(10^{-54}\right)$, so it will precipitate first.

Standard 11
Chemistry

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.