Question
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?
not sure how to go about this
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?
not sure how to go about this
Answers
DrRebel
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
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
DrRebel
Correction ... The [Cr^+3] is 0.039M & the [IO3^-] = 3x = 3(0.039M) = 0.117M. Sorry bout that.