Calculate solubility ( moldm ⁻³ ) of a sparingly soluble electrolyte AB at 298 K if its solubility product is 1.6 10⁻⁵ ?

Step-by-step solution

aligned & For AB , & AB _ ( s ) A _ ( aq ) ⁺+ B _ ( aq ) ⁻ & Here ⁻ x=1, y =1 & ~K _ sp =x^x y ^ y S ^ x+ y =(1)^1(1)^1 ~S ¹⁺¹= S ^2 aligned aligned S & = K _ sp & = 1.6 10⁻⁵ = 16 10⁻⁶ & =4.0 10⁻³ ~mol dm ⁻³ aligned