Question
the solubility product of barium sulphate at 18°c is 1.5x10^-9.its solubility at same temperature is
Answers
bobpursley
write the dissociation eq
BaSO4>>>Ba++ + SO4--
Kc=concBa++ * concSO4--
kc=x^2
so x (solubility)=sqrtKc in moles/liter
now convert that to grams/100ml
x= sqrt (1.5E-9)
to nvert it to solubiliyt (grams/100ml), then
x=.1 molmassinGrams*sqrt(1.5E-9) per 100 ml
BaSO4>>>Ba++ + SO4--
Kc=concBa++ * concSO4--
kc=x^2
so x (solubility)=sqrtKc in moles/liter
now convert that to grams/100ml
x= sqrt (1.5E-9)
to nvert it to solubiliyt (grams/100ml), then
x=.1 molmassinGrams*sqrt(1.5E-9) per 100 ml