Ksp of MCl=1×10−10. Let the molar solubility of MCl in 0.1M‌NaCl be 's mol L'i. ‌MCl→
M+
+Cl−
S
S+0.1
The concentration of Cl−will be (S+0.1)‌mol‌L−1, as 0.1‌mol‌L−1 are provided by 0.1M‌NaCl. Ksp=[M+][Cl−] Ksp=S×(S+0.1)=1×10−10 As the Ksp is very small, the S≪0.1 and therefore can be ignored. S×0.1=1×10−10⇒S=10−9‌mol‌L−1
