Question

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A solution of calcium carbonate is initially supersaturated with Ca++ and CO3= ions, such that the concentration of each is 1.35 x 10-4 mol/L. When equilibrium is finally reached, what is the concentration of calcium ion. Note that if you use the 4-step approach to equilibrium balance shown in class, electro-neutrality need only consider the calcium and carbonate ions, because they are initially in charge balance (i.e. { Ca++}0 = { CO3=}0 and this implies that [Ca++]0 = [CO3=]0 ).

Answer 1

Initially:

[Ca+2] = 1.35*10^-4 M

[CO3-2] = 1.35*10^-4 M

The equilibirum reaction we are talking about is the "solubility" of CaCO3(s)

CaCO3(s) <--> Ca+2 + CO3-2(aq)

Ksp = [Ca+2][CO3-2]

Ksp for CaCO3 = 3.36×10–9

then

Ksp = [Ca+2][CO3-2]

3.36*10^-9 = S*S

S = sqrt(3.36*10^-9) = 5.7965*10^-5

then, this is the actual vlaue which Ca+2 and CO3-2 will have after it preciptiates

recall that initially:

[CO3-2]  = 1.35*10^-4 M

in equilbirium

[CO3-2]  = 1.35*10^-4 - x

and we know

[CO3-2] =  5.7965*10^-5 (in equilbirium

so

1.35*10^-4 - x=  5.7965*10^-5

x = 1.35*10^-4-5.7965*10^-5

x = 0.000077035 M

That is the amount of moles per liter that precipitated

The Calcium ion:

[Ca+2] = 5.7965*10^-5 M