Help me answer this please. After the 2.00 g of basic copper (II) carbonate is p

Question

Help me answer this please. After the 2.00 g of basic copper (II) carbonate is precipitated as described in equation 1, the precipitate will be washed three times with 30.0 ml of deionized water to remove the sodium sulfate and any unreacted sodium carbonate. We know from using the ksp ( equation 2) that each of these washes will remove some of the basic copper (II) carbonate from the final yield. Calculate the final yield of
CU2(OH)2CO3(s) after three rinses to the precision of your scale (0.00).

Explanation / Answer

We know that

Ksp = [Cu+2]^2 [OH-]^2 [CO3-2]

so if solubility of salt = s

Then
Ksp = (2s)2 X (2s)2 X (s)

3 X 10^-12 = 16s^5

0.1875 X 10^-12 = s^5

187.5 X 10^-15 = s^5

so solubility = 2.85 X 10^-3

Solubility = mass of solute / 100grams of water

Initial mass of solute = 2grams

solubility = 2.85 X 10^-3 grams / 100 grams

so solubility in 30 grams = 0.855 X 10^-3 grams

Amount left = 2 - 0.000855 = 1.999145 grams

After second wash and third wash the amount left =2-3X0.000855 = 1.997 grams ( final yield)

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