The chief compound in marble is CaCO3. Marble has been widely used for statues a
ID: 965708 • Letter: T
Question
The chief compound in marble is CaCO3. Marble has been widely used for statues and ornamental work on buildings, including such structures as the Taj Mahal. However, marble is readily attacked by acids via the following reaction.
CaCO3(s)+H+(aq)Ca2+(aq)+HCO3(aq)
Equilibrium constants at 25 C are listed in the table below.
What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in normal rainwater, for which pH=4.20?
Substance Kc Value of Kc CaCO3 Ksp 4.5×109 H2CO3 Ka1 4.3×107 Ka2 5.6×1011Explanation / Answer
We knwo that Ksp of CaCO3 can be represented as
CaCO3 (s) ---> Ca^2+ (aq) + CO3^2- (aq).....Ksp= 4.5 X 10^-9.......(1)
The first dissociation of H2CO3 will be:
H2CO3 + H2O (l) ---> HCO3- (aq) + H+ (aq).......Ka1= 4.3 X 10^-7........(2)
The second dissociation constant
HCO3- (aq) + H2O (l) ---> CO3^2- (aq) + H+ (aq).......Ka2= 5.6 X 10^-11....(3)
We want to get the equation
CaCO3(s)+H+(aq) -->Ca2+(aq)+HCO3(aq) Ksol = Ksp / Ka2 = 80.35
that is we will substract the equation 3 from equation 1
CaCO3(s)+H+(aq) -->Ca2+(aq)+HCO3(aq)
initial a 0 0
Change -x x x
Equilibrium a-x x x
Ksol = [Ca+2] [HCO3-] / [H+]
Ksol = [x]^2 / (a-x)
Given pH = 4.2
pH = -log [H+]
Taking antilog
[H+] = 10^-pH
[H+] = 6.31 X 10^-5
80.35 = [x]^2 / [6.31 X 10^-5-x]
We can ignore x in the denominator
[x]^2 = 80.35 X 6.31 X 10^-5 = 507 X 10^-5
x = 7.12 X 10^-2 M
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