the solubility product of barium sulphate at 18°c is 1.5x10^-9.its solubility at same temperature is
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
To determine the solubility of barium sulphate at 18°C, we need to make use of the solubility product constant (Ksp) for barium sulphate. The solubility product constant is the product of the concentrations of the ions in a saturated solution of the compound.
The equation for the dissolution of barium sulphate (BaSO4) can be represented as follows:
BaSO4 (s) ↔ Ba2+ (aq) + SO42- (aq)
The solubility product expression for barium sulphate is:
Ksp = [Ba2+] * [SO42-]
Given that the solubility product constant (Ksp) of barium sulphate at 18°C is 1.5x10^-9, we can use this value to find the solubility of barium sulphate.
Since barium sulphate dissociates into one barium ion (Ba2+) and one sulphate ion (SO42-) when it dissolves, the concentrations of both ions are the same as the solubility of barium sulphate.
Let's assume the solubility of barium sulphate is 's'. Therefore, [Ba2+] = s and [SO42-] = s.
Substituting these values into the solubility product expression:
Ksp = [Ba2+] * [SO42-]
1.5x10^-9 = (s) * (s)
1.5x10^-9 = s^2
Now, we can solve for 's' by taking the square root of both sides of the equation:
√(1.5x10^-9) = √(s^2)
s = √(1.5x10^-9)
Evaluating this expression, we find that the solubility of barium sulphate at 18°C is approximately 1.22x10^-5 M.