solution is saturated. Example . Indicate find out whether lead chloride gets precipitated or not when mL of .1M lead nitrate and . mL of .
M NaCl solution are mixed? K sp of PbCl is . - PbCl Pb (aq)+2Cl (aq) Ionic product = [Pb ] (s) H O + + ⇀ ↽ [Cl ] - Total volume = . mL Pb NO Pb +2NO M + M ( ⇀ ↽ No of moles of Pb + = Molarity × volume of the solution in litre = .
× × – = – [Pb number of moles of Pb Volume of the solution in L + + ] = mL = . M - NaCl Na Cl M M M → No of moles of Cl – = . × . × – = – [Cl moles L = .
M - - - ] Ionic product = ( . = . - - )( . Since, the ionic product .
- is greater than the solubility product ( . - ) , PbCl will get precipitated. XII U8-Ionic XII U8-Ionic - - - - . .
Determination of solubility product from molar solubility Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution. For a solute X Y , m n X Y (s) mX (aq) +nY (aq) m n n+ m - From the above stoichiometrically balanced equation we have come to know that mole of X Y (s) m n dissociated to furnish ‘m’ moles of X n+ and ‘n’ moles of Y m - if ‘s’ is molar solubility of X Y , m n then [X ]=ms and [ Y ]=ns K = [X ] [Y ] K =( n+ m- sp n+ m m- n sp ∴ ms) (ns) K =(m) (n) (s) m n sp m