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. If the addition of 10.0 mL of a NaIO3 solution to the Cr (III) ion solution causes precipitation to begin, what is the concentration of the NaIO3?
a. 13.9 mol L–1
b. 0.10 mol L–1
c. 1.35 mol L–1
d. 6.25 x 10–2 mol L–1
Ksp = 5E-6 = (Cr^3+)(IO3^-)^3
(Cr^3+) = 0.005 x 125/135 - 4.63E-3
(IO3^-) = cuberoot(5E-6/4.63E-3) = approx 0.1 but you need to confirm that.
So the original concn of the IO3^-, since it was diluted from 10 mL to 135 mL is
0.1 x 135 mL/10 mL) = 1.35 M = (NaIO). Check out all of this; it's late.
To solve this problem, we need to determine the concentration of the NaIO3 solution that causes precipitation of chromium(III) iodate to begin. We can start by writing the balanced chemical equation for the reaction between Cr(III) ion and IO3- ion:
2 Cr3+ + 3 IO3- → Cr2(IO3)6
From the equation, we can see that the mole ratio between Cr3+ and IO3- is 2:3.
Given that the volume of the Cr(III) ion solution is 125 mL and the concentration is 0.0050 mol L-1, we can calculate the moles of Cr3+:
moles of Cr3+ = concentration × volume
= 0.0050 mol L-1 × 0.125 L
= 0.000625 mol
Since the mole ratio between Cr3+ and IO3- is 2:3, the moles of IO3- required for precipitation to begin is:
moles of IO3- = (2/3) × moles of Cr3+
= (2/3) × 0.000625 mol
= 0.0004167 mol
Now, let's find the volume of the NaIO3 solution that contains 0.0004167 mol of IO3- ions. We can use the equation:
moles of solute = concentration × volume
Given that the volume of the NaIO3 solution added is 10.0 mL, we can rearrange the equation to solve for the concentration of NaIO3:
concentration of NaIO3 = moles of solute / volume
= 0.0004167 mol / 0.010 L
= 0.04167 mol L-1
Therefore, the concentration of the NaIO3 solution that causes precipitation to begin is approximately 0.04167 mol L-1.
None of the answer choices provided exactly matches this concentration, but option (d) 6.25 x 10-2 mol L-1 is the closest.