(for $AgCl$ માટે $K_{sp}$ $= 1.8 \times 10^{-10},$ for $PbCl_2$ માટે $ K_{sp}$ $= 1.7 \times 10^{-5}$)
\(\left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{10^{-1}}=1.8 \times 10^{-9} \,\mathrm{M}\)
\(K_{s p}\left[\mathrm{PbCl}_{2}\right]=\left[\mathrm{Pb}^{2+}\right][\mathrm{Cl}]^{2}\)
\(\left[\mathrm{Pb}^{2+}\right]=\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}}=1.7 \times 10^{-3} \,\mathrm{M}\)
(આપેલ : $K _{ b }\left( NH _4 OH \right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11=$ $1.04)$