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

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

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

To find the concentration of Cr(III) ion remaining in solution after adding 75.0 mL of NaIO3, we will use the concept of common ion effect and the Ksp value of chromium (III) iodate.

Let's break down the steps:

Step 1: Write the balanced equation for the dissociation of chromium (III) iodate:
Cr(IO3)3(s) ⇌ Cr3+(aq) + 3 IO3-(aq)

Step 2: Calculate the initial concentration of Cr3+ ions:
Given that the initial solution contains 0.0050 mol L-1 Cr (III) ion and the volume is 125 mL, we can calculate the initial moles of Cr3+ ion:
Initial moles of Cr3+ = initial concentration x volume (in L)
= 0.0050 mol L-1 x (125 mL / 1000 mL/L)
= 0.000625 mol

Step 3: Apply the common ion effect:
The common ion effect states that the solubility of a salt decreases when a common ion is added to the solution. In this case, adding NaIO3 solution will introduce additional IO3- ions, which will shift the dissociation equilibrium to the left, reducing the concentration of Cr3+ ions remaining.

Step 4: Calculate the moles of Cr3+ ions that have precipitated:
Since the molar ratio between Cr(IO3)3 and Cr3+ is 1:1, the moles of Cr3+ that have precipitated are equal to the initial moles of Cr3+ minus the remaining moles of Cr3+:
Moles of Cr3+ precipitated = initial moles of Cr3+ - remaining moles of Cr3+
= 0.000625 mol - X (remaining moles of Cr3+)

Step 5: Calculate the concentration of the remaining Cr3+ ion:
To find the concentration, divide the remaining moles of Cr3+ by the final volume of the solution.

Final volume = initial volume + volume of NaIO3 added
= 125 mL + 75 mL
= 200 mL

Concentration of the remaining Cr3+ ion = remaining moles of Cr3+ / final volume (in L)

Step 6: Calculate the remaining moles of Cr3+ ion:
To find the remaining moles of Cr3+ ion, we need to find the initial moles of IO3- ions added and use the balanced equation to determine the moles of Cr3+ ions that have precipitated.

Moles of IO3- ions added = concentration x volume (in L)
= 0.0050 mol L-1 x (75 mL / 1000 mL/L)
= 0.000375 mol

Since the stoichiometric ratio is 1:3 between Cr3+ and IO3-, the moles of Cr3+ ions that have precipitated will be three times the moles of IO3- ions added.

Moles of Cr3+ ions that have precipitated = 3 x moles of IO3- ions added
= 3 x 0.000375 mol
= 0.001125 mol

Step 7: Calculate the remaining moles of Cr3+ ion:
remaining moles of Cr3+ = initial moles of Cr3+ - moles of Cr3+ ions that have precipitated
= 0.000625 mol - 0.001125 mol
= -0.0005 mol (negative value indicates that all Cr3+ ions have precipitated)

Step 8: Calculate the concentration of the remaining Cr3+ ion:
Concentration of the remaining Cr3+ ion = remaining moles of Cr3+ / final volume (in L)
= -0.0005 mol / 0.2 L
= -0.0025 mol L-1

The concentration of the remaining Cr(III) ion in solution is -0.0025 mol L-1. Since concentrations cannot be negative, it means that all Cr(III) ions have precipitated as chromium (III) iodate.