Calculate the molar solubility of in the following substances. Pure water and 1.5×10−3 K2CrO4
Your question makes no sense to me.
1. 1.5E-3 WHAT?
2. Do you want the molarity calculated?
3. If you refer to the solubility, it is completely soluble.
To calculate the molar solubility of a substance, we need to know its solubility product constant (Ksp) and the conditions under which the solubility is being measured.
In this case, we are given the solubility product constant (Ksp) for K2CrO4, which is 1.5 × 10^(-3). The Ksp expression for K2CrO4 is:
Ksp = [K+]^2[CRO4^2-]
To find the molar solubility of K2CrO4 in pure water, we assume that the concentration of K2CrO4 is x M. Since K2CrO4 dissociates into two K+ ions and one CrO4^2- ion, the concentration of K+ ions and CrO4^2- ions will both be 2x M.
Using the solubility product expression, we can substitute these values into the equation:
1.5 × 10^(-3) = (2x)^2 * (2x)
1.5 × 10^(-3) = 4x^3
x^3 = (1.5 × 10^(-3)) / 4
x^3 = 3.75 × 10^(-4)
Taking the cube root of both sides:
x = (3.75 × 10^(-4))^(1/3)
x ≈ 0.061 M
Therefore, the molar solubility of K2CrO4 in pure water is approximately 0.061 M.
Note: The molar solubility may change in different solvents or under different conditions. This calculation assumes an ideal solution and dilute conditions.