determine solublityof

• i) ferric hydroxide when Ksp= 1 * 10^-38
• ii) lead chloride when Ksp= 1.6* *10^-5

Dear Student,

i)
The computation are expressed below,

$$\begin{gathered} Fe(OH{)}_3\rightleftharpoons F{e}^{3+}+3 O{H}^{-} \\ {K}_{sp}=[F{e}^{3+}\left]\right[O{H}^{-}{]}^3 \\ 1.1\times {10}^{-36}=\left(x\right)(x{)}^3 \\ x=1\times {10}^{-9} \raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right. \\ for conversion molar mass is used, \\ F{e}^{3+}={10}^{-9} \raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right. \times 55.85 \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.=5.58\times {10}^{-10} \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$L$}\right. \\ O{H}^{-}={10}^{-9} \raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right. \times 51 \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. = 5.4\times {10}^{-10} \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$L$}\right. \end{gathered}$$

N:B:- Check Ksp value for 1 x 10^-38

ii)

PbCl₂  →  Pb⁺⁺   +   2Cl¯

   s   2s

  K_sp = [Pb⁺⁺][ Cl¯]²

  K_sp = s x 4s² = 4s³

   1.6 x 10^-5 = 4s³

   16 x 10^-6 = 4s³

   s = 4 x 10^-2  

   s = 1.5874 x 10^-2 M = solubility

Regards,