AP EAMCET202021 Sep 2020Morning ShiftChemistryIonic EquilibriumActual
The solubility of ( AgBr ) with solubility product (5.0 10⁻¹³ ) at (298 ~K ) in (0.1 M NaBr ) solution would be
Options
- A(7 10⁻⁶ M )
- B(5 10⁻¹² M )
- C(5 10⁻¹⁴ M )
- D(5 10⁻⁶ M )
Correct answer
B. (5 10⁻¹² M )
Step-by-step solution
Let, the solubility of ( AgBr ) be (S ~mol / L ). ( AgBr Ag ⁺+ Br ⁻ ) Hence, ( [ Ag ⁺ ] [ Br ⁻ ]=5 10⁻¹³ ) Given that, ( [ Br ⁻ ]=0.1 ) (from ( NaBr )) So, ( [ Ag ⁺ ]= (5 10⁻¹³ ) / 0.1=5 10⁻¹² M ) It means solubility in ( NaBr ) is (5 10⁻¹² ). Hence, option (b) is correct.