Question

How many grams of AgBr will dissolve in 1.0 L of water containing a Br- concentration of 0.050 M? Ksp= 5.0 * 10^-13

Answer 1

1. Calculate solubility of AgBr in 1.0 L of water containing Br^- concentration = 0.05 M ,

using solubility product principle as,

...................Ag Br (aq) <==========> Ag⁺ (aq) + Br ^- (aq)

......................... s...........................................s ..............(s + 0.05)......

( where s represents solubility of AgBr in moles per litre )

The equilibrium concentrations in terms of s moles / L are

................................[ Ag⁺ ] = s

................................[ Br ^- ] = ( s + 0.05 )

......................................K_sp   = [ Ag⁺ ] [ Br - ]

therefore plugging in the above values -     

...........................(s) (s + 0.05 ) = 5.0 x 10^-13  

...........................s²  + 0.05 s = 5.0 x 10^-13

.......( Since .s is very small as compared to Br ^- concentration , the factor s² can be ignored )

......................................... ....s = ( 5.0 x 10^-13 ) / 0.05

.....................................................= 1.00 x 10^-11 M

and weight of AgBr dissolved in 1.0 L of water = No. of moles of AgBr per Litre x Molecular mass

............................................................................. = ( 1.00 x 10^-11 ) ( 187.7722 )

............................................................................. = 1.87 x 10 ^-9 gms.