Chromium (III) iodate has a Ksp at 25°C of 5.0 x 10–6.
125 mL of a solution contains 0.0050 mol L–1 Cr (III) ion.
Possible answers:
a.
8.4 x 10–2 mol L–1
b.
2.2 x 10–3 mol L–1
c.
6.3 x 10–2 mol L–1
d.
2.1 x 10–2 mol L–1
What is the molar solubility of chromium (III) iodate at 25°C?
To determine the molar solubility of chromium (III) iodate at 25°C, we need to use the concept of solubility product constant (Ksp).
The Ksp for chromium (III) iodate is given as 5.0 x 10–6 at 25°C, which represents the equilibrium expression:
Ksp = [Cr(III) ion] * [IO3-]
We are given the concentration of Cr(III) ion in solution as 0.0050 mol L–1.
To calculate the molar solubility, we need to assume that all of the Cr(III) ion in solution comes from the dissociation of chromium (III) iodate.
Let's assume the molar solubility as "x" mol L–1. As chromium (III) iodate dissociates in a 1:3 ratio of Cr(III) ion to IO3-, the concentration of Cr(III) ion would also be "3x" mol L–1.
Substituting these values into the equilibrium expression for Ksp:
5.0 x 10–6 = (3x) * x
5.0 x 10–6 = 3x^2
Rearranging the equation, we have:
3x^2 = 5.0 x 10–6
Dividing both sides by 3:
x^2 = (5.0 x 10–6) / 3
x^2 = 1.67 x 10–6
Taking the square root of both sides:
x ≈ 1.29 x 10–3 mol L–1
Therefore, the molar solubility of chromium (III) iodate at 25°C is approximately 1.29 x 10–3 mol L–1.
Among the given options, the closest answer is:
b. 2.2 x 10–3 mol L–1
Please note that the actual solubility of a compound can be affected by factors such as temperature and other chemical species present in the solution, so the calculated value may not be exact in practice.