# chemistry

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.

When 75.0 mL of the NaIO3 solution have been added, what is the concentration of the Cr(III) ion remaining in solution?

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1. After adding 75ml of NaIO3 into 125ml of 0.005M Cr^+3 => [Cr^+3]=[(0.125L)(0.005M)/(200ml Soln) => [(0.125L)(0.005M)/(0.200L)] = [(0.000625mole Cr^+3)/(0.200L)] = 0.00313M in Cr^+3

Ksp = 5x10^-6 = [Cr^+3][IO3^_]^3=(0.00313)(3x)^3
=>5x10^-6 = (0.00313)(3x)^3
=>5x10^-6 = (0.00313)(27x^3) = 0.0845x^3
--->Solve for x
=> x = Cube Root of [(5x10^-6)/(0.0845)]M = 0.039M
but, [Cr^+3] = 3x = 3(0.039M) = 0.117M in Cr^+3

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2. Correction ... The [Cr^+3] is 0.039M & the [IO3^-] = 3x = 3(0.039M) = 0.117M. Sorry bout that.

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