Final answer:
The molar solubility of BaCrO4 in pure water can be calculated by taking the square root of its Ksp value which is 2.1 x 10−10. The molar solubility is found to be 1.45 x 10−5 M.
Step-by-step explanation:
To find the molar solubility of BaCrO4, we'll use the solubility product constant (Ksp). The dissociation of BaCrO4 in water is represented as BaCrO4 (s) ⇌ Ba2+ (aq) + CrO42- (aq). As it dissolves, it forms equimolar concentrations of Ba2+ and CrO42-. If we let the molar solubility be 's', then the concentration of Ba2+ and CrO42- at equilibrium will both be 's'. The Ksp expression is Ksp = [Ba2+][CrO42-] = s * s = s2. With the given Ksp of 2.1 x 10−10, we can solve for 's'.
Setting up the equation: Ksp = s2, we get 2.1 x 10−10 = s2. To find 's', we take the square root of the Ksp: s = √(2.1 x 10−10)
s = 1.45 x 10−5 M
Therefore, the molar solubility of BaCrO4 in pure water is 1.45 x 10−5 M.