Shells of mussels (clams) are mainly of calcium carbonate. Usually the shell is

Question

Shells of mussels (clams) are mainly of calcium carbonate. Usually the shell is stable and its content is in equilibrium with the surrounding. Mussels need to built up or reduce their shell, however, for additional surrounding.

Assume that the concentration of CA2+ ions in seawater and in the vicinity of a mussel bank is

(1.0 X 10^-2) M. Also, assume that the total concentration of "inorganic carbon" in seawater (this is carbon in chemical species with carbonate entities is (2.3 X 10^-3) M

Calculate the pH (correct to two decimals) of the water in the vicinity of mussels so that their shell remains stable.

Ks(CaCO3)=9.9x10^-9 M^2 ; Ka1=4.3x10^-7 M ; Ka2=4.8x10^-11 M

Explanation / Answer

From Ksp value we can calcualte concentration of CO3-2

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

[Ca+2] = 0.01 M

[CO3-2] = (Ksp)/ [Ca+2]

[CO3-2] = (9.9x10^-9) / 0.01 = 9.9 X 10-7

2.3 X 10^-3 = [HCO3-] + [ CO3-2 ] + [H2CO3]

[HCO3-] + [H2CO3] = 2.3 X 10^-3 [the concentration of carbonate ions is negligible as compared to total inorganic carbon concentration] ......................(1)

Now,

H2CO3 ---> H+ + HCO3-

Ka1 = [HCO3-][H+] / [H2CO3] = [HCO3-]2 / [H2CO3]

4.3x10-7 X[H2CO3] = [HCO3-]2 ............... (2)

Equating (1) and (2)

[HCO3-]2 /4.3x10-7 + [H2CO3] = 2.3 X 10^-3

[HCO3-]2 + [H2CO3] X 4.3x10-7 - 2.3 X 10^-3 X4.3x10-7 = 0

On solving

[HCO3-] = 0.0000312 = 3.12 X 10^-5 = [H+]

HCO3- ---> H+ + CO3-2

Ka2 = [H+][CO3-2] / [ HCO3- ]

[H+] due to dissociation of bicarbonate ions = (Ka2 X [ HCO3- ] )1/2 = (4.8 X 10-11X 3.12 X 10^-5)1/2 = 3.87 X 10^-8 M

total [H+] =3.12 X 10^-5 +3.87 X 10^-8 M = 3.124 X 10^-5 M

pH = -log[H+] = 4.5

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