Question

Given that, for AgCl, at 25°C, Ksp 1.6x10 10 From a 1.0x102-M NaCl solution, precipitation of AgCl just begins at what concentration of added Ag*? I Tries 0/5 Submit Answer

Answer 1

Ag^{​​​​​+}+Cl^{​​​​​-} = AgCl

Ionic product (Q) of AgCl =[Ag_​​​^​​​+][Cl^{​​​​-}]

Concect of solubility if K_{​​​​​​sp} value is <Q value then it is precipitated .

If Q value = K_{​​​​​sp}​​​​ then it is just precipitated.

Here in the problem [Cl^-] is come from NaCl.=1.0×10^-2

[Ag^​​​+] =?

Given also K_{​​​​​​sp} of AgCl =1.6×10^-10

According to the given rule

Q= K_sp

>K_sp =[Ag^​​​+] [Cl^{​​​​​-}​]

[Ag⁺] =K_sp​/Cl^{​​​​​-}

= 1.6×10^-10/1.0×10^-2

=1.6×10^-8 M