In: Chemistry
Write the balanced chemical equation for the dissolving of copper carbonate in water, and write its equilibrium expression. Calculate the concentration of copper in the saturated solution using the experimental cell potential (0.21 volts). Calculate the Ksp for copper (II) carbonate using the concentration of copper found in the previous question (the previous question is "calculate the concentration of copper...").
Case1:
Balanced chemical equation for the dissolving of copper
carbonate in water
CuCO3 (s) <=> Cu+2 (aq) +
CO3-2 (aq) ------(1)
case 2:
The equilibrium expression (Ksp) for the reaction 1
Ksp = [Cu+2] [CO3-2]
Case 3:
Using the experimental cell potential ( .21V), calculate the
concentration of copper in the saturated solution
this experiment usually
compare the [Cu+2] level in the beaker with CuCO3, with
the [Cu+2] level in a 1 molar solution. it is called a
concentration cell
copper's Standard Reduction Potential of Eo= +
0.340V for this reaction
Cu2+ + 2 e − → Cu
while the reverse, copper's Standard Oxidation Potential of Eo= - 0.340V for this reaction
Cu →Cu2+ + 2 e −
for the reaction:
Cu2+ + 2 e − → Cu
the standard voltage for a concentration cell is zero
but because the CuCO3 lowers the [Cu+2], that
cell will have an EMF of a lower voltage by 0.21 v
Nernst equation is given by :
E = Eo - (0.0592 / n) (log Q)
0.21 = zero - (0.0592 / 2 mol e- )* log (prod /
reactants)
0.21 = zero - (0.0296)*log ([Cu+2] / [Cu+2])
for the cell to work, i,e, to proceed ---> --->
the higher "usually 1Molar Cu+2" must be the reactant
0.21 = - (0.0296) (log [Cu+2] / [1M])
0.21 = - (0.0296) (log [Cu+2] )
0.21 / - (0.0296) = (log [Cu+2] )
log [Cu+2] = -7.0946
[Cu+2] = 8.0430 e-8 Molar
Case 4:
Using the answer from (case 3), calculate the value of Ksp for
copper (II) carbonate
you will have to use your lab's concentration of carbonate...
usuall it is 1 Molar, having added a few drops of copper II
solution to a 1 Molar Na2CO3 solution
if that is the case for your exp...
then
Ksp = [Cu+2] [CO3-2]
Ksp = [8.043 e-8 M] [1M]