Question

at 298k temperature, the solubility of AgCl is 1.6×10-3 g/L. Calculate the solubility product of AgCl at this temperature.

Answer 1

SoluTion :-

The solubility equilibrium in the saturated solution is :

$\sf {AgCI\ (s) \overset{aq}{\Leftrightarrow} Ag+(aq)+CI^{-}(aq) }$

The solubility of AgCl is 1.06 × 10⁻⁵ mole per litre.

$\sf {[Ag^{+}(aq)]=1.06 \times 10^{-5}molL^{-1} }\\\\\\\sf {[Cl^{-}(aq)]=1.06\times 10^{-5}molL^{-1}}\\\\\\\sf {K_{sp}=[Ag^{+}(aq)][Cl^{-}(aq)] }\\\\\\\sf {=(1.06 \times 10^{-5}molL^{-1}) \times (1.06 \times 10^{-5}molL^{-1}) }\\\\\sf {=1.12 \times 10^{-10} mol^{2}L^{-2}}$