17.99 The solubility of CaCO, is pH dependent. (a) Calculate the molar solubilit
ID: 1036791 • Letter: 1
Question
17.99 The solubility of CaCO, is pH dependent. (a) Calculate the molar solubility of CaCO, (Kp 4.5 X 109) neglecting the acid-base character of the carbonate ion. (b) Use the Kb expression for the CO, ion to determine the equilibrium constant for the reaction CaCO3(s) + H2O(1) (c) If we assume that the only sources of Ca2+,HC03, and OH ions are from the dissolution of CaCOs, what is the mo- lar solubility of CaCO, using the equilibrium expression from part (b)? (d) What is the molar solubility of CaCO; at the pH of the ocean (8.3)? (e) If the pH is buffered at 7.5, what is the molar solubility of CaCO3?Explanation / Answer
(a) The dissociation of CaCO3 is given as
CaCO3 (s) <=====> Ca+ (aq) + CO32- (aq)
Ksp = [Ca2+][CO32-] ……..(1)
=====> 4.5*10-9 = (x).(x) (1:1 ionization of CaCO3)
=====> x2 = 4.5*10-9
=====> x = 6.708*10-5 ? 6.71*10-5
The molar solubility of CaCO3 in water (ignoring acid-base character of CO32-) is 6.7*10-5 M (ans).
(b) Consider the ionization of H2CO3 as below.
H2CO3 (aq) -------> H+ (aq) + HCO3- (aq); Ka1 = 4.3*10-7
HCO3- (aq) --------> H+ (aq) + CO32- (aq); Ka2 = 4.8*10-11
Given CO32- participates in acid-base reaction, we can write the reaction as
CO32- (aq) + H2O (l) --------> HCO3- (aq) + OH- (aq)
Kb = Kw/Ka2 = [HCO3-][OH-]/[CO32-]
=====> Kb = [HCO3-][OH-]/[CO32-] = (1.0*10-14)/(4.8*10-11) = 2.08*10-4 ? 2.1*10-4 ……(2)
The equilibrium constant for the given reaction is given as
K = [Ca2+][HCO3-][OH-] = ([Ca2+][CO32-])*([HCO3-][OH-]/[CO32-])
= Ksp*Kb = (4.5*10-9)*(2.1*10-4) = 9.45*10-13 (ans).
(c) Let S be the solubility of CaCO3 in water (consider acid-base character of CO32- now).
CaCO3 (s) + H2O (l) <=====> Ca2+ (aq) + CO32- (aq) + OH- (aq)
S S S
We have
Ksp = [Ca2+][CO32-][OH-]
=====> 4.5*10-9 = S*S*S = S3
=====> S = 1.65*10-3
The molar solubility of CaCO3 (consider the acid-base character of CO32-) is 1.65*10-3 M (ans).
(d) The pH of ocean water is pH = 8.3; therefore,
pOH = 14 – pH = 14 – 8.3 = 5.7
Therefore, [OH-] = antilog (-5.7) = 1.99*10-6
Let S be the solubility of CaCO3 in water; therefore, we have
Ksp = S*(S – x) where x = concentration of CO32- reacted with H2O ……(3)
Again, we have,
CO32- (aq) + H2O (l) --------> HCO3- (aq) + OH- (aq)
(S – x) x 1.99*10-6
Kb = (x).(1.99*10-6)/(S – x)
=====> 2.1*10-4 = (x).(1.99*10-6)/(S – x)
=====> x/(S – x) = 2.1*10-4/(1.99*10-6) = 105.528
=====> x = 105.528*(S – x)
=====> x = 105.528S – 105.528x
=====> 106.528x = 105.528*S
=====> x = 105.528/106.528*S = 0.99S …….(3)
Put in (3) and get
4.5*10-9 = S*(S – x) = S*(S – 0.99S) = S*(0.01*S)
=====> 4.5*10-9 = 0.01*S2
=====> S2 = 4.5*10-9/0.01 = 4.5*10-7
=====> S = 6.71*10-4
The molar solubility of CaCO3 in ocean water is 6.71*10-4 M (ans).
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