Question

Consider a solution that is 1.34 x 10^-5 M in Ag(NO3). Which of the following concentrations of KCl will result in the precipitation of AgCl? (Ksp = 1.8 x 10^-10)

Please show work and all steps.

Answer 1

The reaction will occur as

AgNO₃ (aq) + KCl (aq) $\rightleftharpoons$ AgCl (s) + KNO₃ (aq)

Concentration of solution containing AgNO₃ = 1.34 x 10^-5 M

AgNO₃ (aq)      $\rightleftharpoons$                  Ag⁺        +       NO₃              

In case of precipitation of AgCl,

AgCl (s)         $\rightleftharpoons$             Ag⁺ +Cl^-

Both the cation and anion has the same 1 charge so need to balance the equation

AgCl (s)          $\rightleftharpoons$        Ag⁺    + Cl^-

At initial stage conc.   (1.34 x 10^-5 M ) 0

  (as Ag⁺ is the common ion)

At change conc.                              x x  

At Equillibrium { x + (1.34 x 10^-5 M )} (0 + x)

K_sp =         (Ag⁺ ) (Cl^- )    

K_sp =         { x + (1.34 x 10^-5 M ) x (x)}                                  take x<<<<   (1.34 x 10^-5 M ) so can be neglected

K_sp =           (1.34 x 10^-5 M ) x

K_sp =           1.8 x 10 ^-10         (given)

Hence:                           x = (1.8 x 10^-10 ) / ( 1.34 x 10^-5 )  

                                         X = 1.34 x 10^-5 M ^is concentration of AgCl will result in precipitation.