K sp in above equation at 298K is . × – . This means that for solid barium sulphate in equilibrium with its saturated solution, the product of the concentrations of barium and sulphate ions is equal to its solubility product constant. The concentrations of the two ions will be equal to the molar solubility of the barium sulphate.
If molar solubility is S, then . × – = (S)(S) = S or S = . × – . Thus, molar solubility of barium sulphate will be equal to .
× – mol L – . A salt may give on dissociation two or more than two anions and cations carrying different charges. For example, consider a salt like zirconium phosphate of molecular formula (Zr + ) (PO – ) . It dissociates into zirconium cations of charge + and phosphate anions of charge – .
If the molar solubility of zirconium phosphate is S, then it can be seen from the stoichiometry of the compound that [Zr + ] = 3S and [PO – ] = 4S and K sp = (3S) (4S) = (S) or S = { K sp / ( × )} / = ( K sp / ) / A solid salt of the general formula M X x y q with molar solubility S in equilibrium with its saturated solution may be represented by the equation: M x X y (s) xM p+ (aq) + yX q– (aq) (where x × p + = y × q – ) And its solubility product constant is given by: K sp = [M p+ ] x [X q– ] y = (xS) x (yS) y ( . ) = x x . y y . S (x + y) S (x + y) = K sp / x x .
y y S = ( K sp / x x . y y ) / x + y ( . ) The term K sp in equation is given