The chief compound in marble is CaCO3. Marble has been widely used for statues a

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 (Figure 1) . 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=5.60?

Express your answer to two significant figures and include the appropriate units.

5.6×1011

5.6×1011

Explanation / Answer

We know 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 = 5.6

pH = -log [H+]

Taking antilog
[H+] = 10^-pH
[H+] = 2.51*10^-5

80.35 = [x]^2 / [2.51 X 10^-6-x]

We can ignore x in the denominator
[x]^2 = 80.35 X 2.51 X 10^-6 = 2.107X 10^-4

x = 0.0145 M

= 1.4*10^-2 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.