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.
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
1 answer
1 answer