 
                 
                InterviewSolution
 Saved Bookmarks
    				| 1. | The solubility of AgCl is `1xx10^(-5)mol//L`. Its solubility in 0.1 molar sodium chloride solution isA. `1xx10^(-10)`B. `1xx10^(-5)`C. `1xx10^(-9)`D. `1xx10^(-4)` | 
| Answer» Correct Answer - C `K_(sp)` of AgCl=(solublility of `AgCl)^(2)` `=(1xx10^(-5))^(2)=1xx10^(-10)` Suppose its solubility in 0.1 M NaCl is mol/L. `AgCl Leftrightarrow underset(x)(Ag^(+))+underset(x)(Cl^(-))` `NaCl Leftrightarrow underset(0.1M)(Na^(+))+underset(0.1M)(Cl^(-))` `[Cl^(-)]=(x+0.1)M` `k_(sp)"of AgCl"=[Ag^(+)][Cl^(-)]` `=x xx(x+0,1)` `1xx10^(-10)=x^(2)+0,1x` Higher power of x are neglected `1xx10^(-10)=0.1x` `x=1xx10^(-9)M` | |