G. A solution contains one or more of the following ions: Ba +2 , Cu +2 , and Co

Question

G. A solution contains one or more of the following ions: Ba+2, Cu+2, and Co+2. When you add sodium sulfate to the solution, no precipitate forms. When you add sodium iodide, a dark precipitate forms. You filter off the precipitate and add sodium carbonate to the remaining solution, producing a cloudy precipitate. Which ions were present in the original solution? Write a net ionic equation for the formation of each of the precipitates observed?

H.H. Lets say we have a waste container from our lab containing the following ions: Ba+2, Cu+2, and Ni+2. Based on your findings, suggest a sequence for adding the sodium solutions in such a way that we could separate each of these ions by selective precipitation. Write a net ionic equation for the formation of each precipitate.

I.Have you ever noticed the soapy ring that forms around a dirty tub? This is the product of hard water

containing Ca+2 and Mg+2 ions reacting with soap. This soapy scum could also accumulate on our

clothes if it wasn

Explanation / Answer

1) First reaction shows Cu2+
2Cu2+ + 4I- -------> 2CuI + I2

Second reaction shows Ba2+
Ba2+ + CO3 2- ------> BaCO3

2) We can start by adding sodium sulfate. Of the three cations mentioned, only Barium forms an insoluble sulfate in water.

Ba(+2) + SO4(-2) ---> BaSO4 (s)

If we now filter off this precipitate, we are left with a solution containing just Cu(+2) and Ni(+2). We now add sodium iodide. Nickel iodide is soluble in water, but copper iodide is not.

Cu(+2) + 2 I(-) ---> CuI2 (s)

Copper (II) iodide is not stable, and will be reduced very quikcly to Copper (I) iodide.

2 CuI2 ---> 2 CuI + I2.

We can now filter off this precipitate, and the resulting solution will contain just Ni(+2) cations. This can be precipitated using sodium carbonate.

Ni(+2) + CO3(-2) ---> NiCO3 (s).

3) 20.1 gallons times 3.785 litres per gallon = 76 liters water.

Moles Ca++ = 3.6 x l0^-3 Molar times 76 liters = .27 moles Ca++

Moles Mg++ = l.0 x l0^-3 Molar times 76 Liters = .076 moles Mg++

Each mole of Ca++ combines with one mole of CO3=

and each mole of Mg++ combines with one mole of CO3= so we can add the moles of Ca++ and Mg++ together to get .35 moles of diprotic ions which will combine with the same number of moles of

CO3= ion

so we will need .35 moles of CO3= ion which can be found in .35 moles of Na2CO3 which will

weigh .35 moles times 110 grams per mole or 37 grams Na2CO3

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