the solubility of barium chloride at 298k is 1.05*10 raised -5. calculate the solubility product?
i want the answer please
To calculate the solubility product (Ksp) of barium chloride, we need to understand the chemical equation representing its dissociation in water and how the equilibrium expression is formed.
The chemical equation for the dissociation of barium chloride (BaCl2) in water is:
BaCl2(s) ⇌ Ba2+(aq) + 2Cl-(aq)
The equilibrium expression for this reaction is:
Ksp = [Ba2+][Cl-]^2
Given that the solubility of barium chloride at 298K is 1.05 × 10^-5, we can say that [Ba2+] = 1.05 × 10^-5 M and [Cl-] = 2 × 1.05 × 10^-5 M.
Plugging these values into the equilibrium expression, we get:
Ksp = (1.05 × 10^-5)(2 × 1.05 × 10^-5)^2
Ksp = 2.205 × 10^-15
Therefore, the solubility product (Ksp) of barium chloride at 298K is 2.205 × 10^-15.
Check your post. Surely you don't mean BaCl2. The solubility of BaCl2 is far more than 1.05E-5 and when you say 1.04E-5 is that grams, moles, tons. And what is the volume? mL, L, oceans?
So you want the answer. You don't want to know how to work it?