A saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found to be 0.00072 M. Set up an ICE chart and calculate the equilibrium (IO3-) and the Ksp

0.02644 M = [IO3-]

Ksp = 0.0000005 = 5*10^-7

Explanation:

Step 1: Data given

Asaturated solution of Cu(IO3)2 was prepared in KIO3

Concentration of KIO3 = 0.0250 M

The copper concentration was found to be 0.00072 M

Step 2: Equations

Cu(IO3)2(s) + aq ⇔Cu2+(aq) + 2IO3 -  (aq)

Step 3: The initial concentrations

The initial concentration of KIO3 = 0.0250 M

The initial concentration of Cu2+ is 0 M

Step 4: The changed concentrations

After preparing the Cu2+ in KIO3-; the concentration of Cu2+ will be changed with X

For every 1 Cu2+ we have 2 IO3-

This means if the concentration of Cu2+ is changed by X; the concentration of IO3- is changed by 2X. This gives a total concentration of 0.0250 + 2X for IO3-

Step 4:

Said is that the copper concentration was found to be 0.00072 M

This means the final concentration of Cu2+ is 0.00072 M or also said X = 0.00072

Now we can calculate the final IO3- concentration

0.0250 + 2X  = [IO3-]

0.0250 + 2* 0.00072 = [IO3-]

0.0250 + 0.00144 = [IO3-]

0.02644 M = [IO3-]

Step 5: Calculate Ksp

Cu(IO3)2(s)  ⇔Cu2++ 2IO3 -  

Ksp of Cu(IO3)2 = [Cu+2]*[IO3-]²

Ksp = 0.00072 * 0.02644²

Ksp = 0.0000005 = 5*10^-7

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Explanation:

1. The ICE table

2. Concentration of IO₃⁻

At equilibrium, [Cu²⁺] = 0.000 72 mol·L⁻¹, so x = 0.000 72.

The new ICE table becomes

3. Ksp

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