JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
When silver nitrate solution is added to potassium iodide solution then the sol produced is :
- (a) AgI / I⁻
- (b) AgI / Ag⁺
- (c) KI / NO₃ -
- (d) AgNO₃ / NO₃ -
Answer and solution
Answer (a) AgI / I⁻
Solution
Chemistry ma’am · 4 steps
Ask which ion is in excess: AgNO₃ goes into KI, so iodide is left over and sticks to the AgI.
- What forms. AgNO₃ + KI → AgI + KNO₃. The colloidal particles are AgI; KNO₃ stays dissolved.
- Which ion is excess. Silver nitrate is added to the potassium iodide solution, so at every stage KI is in excess. The common ions left over in the solution are I⁻ ions.
- Preferential adsorption. A precipitate adsorbs the ion it shares with its own lattice. AgI contains I⁻, so it adsorbs the excess I⁻ and the particles become negatively charged.
- So. The sol is AgI/I⁻, a negative sol, which is option (a).
The idea to keep. Whatever is in the beaker first is in excess. The sol particle adsorbs its own common ion from that excess and takes its charge.
Answer from the NTA final answer key. Solution by PrepWiser.