Is #CaCO_3# acidic, basic, or neutral?

2 Answers
Jul 3, 2017

Answer:

pH = 9.7

Explanation:

pH of Saturated #CaCO_3# at #25^oC#:

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Jul 3, 2017

It's basic, due to the amount of #"OH"^(-)# it generates (greater than #10^(-7)# #"M"#). The #"pH"# should be somewhere near #8# at #25^@ "C"# and #"1 atm"#, with normal #"CO"_2# partial pressures in the air.


We can examine the dissociation into water using its #K_(sp)#:

#"CaCO"_3(s) rightleftharpoons "Ca"^(2+)(aq) + "CO"_3^(2-)(aq)#

#K_(sp) = ["Ca"^(2+)]["CO"_3^(2-)] = 3.3 xx 10^(-9)#

giving each concentration in a saturated solution at #25^@ "C"# and #"1 atm"# as

#["Ca"^(2+)] = ["CO"_3^(2-)] = sqrt(K_(sp))#

#= 5.74 xx 10^(-5)# #"M"#

Considering the base's association in water, we obtain the #K_b# of #"CO"_3^(2-)# from the #K_a# of #"HCO"_3^(-)# at #25^@ "C"# and #"1 atm"#:

#K_b("CO"_3^(2-)) = K_w/(K_a("HCO"_3^(-)))#

#= (1 xx 10^(-14))/(4.8 xx 10^(-11)) = 2.08 xx 10^(-4)#

Due to the large #K_b#, we expect most of the #"CO"_3^(2-)# to be gone. The equilibrium association process is

#"CO"_3^(2-)(aq) + "H"_2"O"(l) rightleftharpoons "HCO"_3^(-)(aq) + "OH"^(-)(aq)#

with the mass action expression

#K_b = (["OH"^(-)]["HCO"_3^(-)])/(["CO"_3^(2-)])#

#= x^2/(5.74 xx 10^(-5) - x) = 2.08 xx 10^(-4)#

Due to the small concentration and the not small #K_b#, we cannot make the small #x# approximation on this, as the less of it there is, the more of the #"CO"_3^(2-)# is going to associate.

Solving for the quadratic equation form gives

#2.08 xx 10^(-4)(5.74 xx 10^(-5)) - 2.08 xx 10^(-4) x - x^2 = 0#

This gives rise to two solutions, and the physical solution has

#color(blue)(x = 4.68 xx 10^(-5))# #color(blue)("M")#,

as the (first) equilibrium concentration of #"OH"^(-)#, larger than #10^(-7) "M"# (the requirement for basicity at #25^@ "C"# and #"1 atm"#).

This gives a preliminary #"pH"# of #9.67#, but this is not entirely correct. There are two opposing equilibria for #"OH"^(-)# here:

Forward Reaction Forming #bb("OH"^(-))#

#"CO"_3^(2-)(aq) + "H"_2"O"(l) -> "HCO"_3^(-)(aq) + "OH"^(-)(aq)#

Backwards Reaction Forming #bb("OH"^(-))#

#"Ca"^(2+)(aq) + 2"OH"^(-)(aq) larr "Ca"("OH")_2(s)#

Now, in principle, this #"OH"^(-)# would react with the remaining #"Ca"^(2+)# to form #"Ca"("OH")_2#. That #K_(sp)#, however, is actually #5.5 xx 10^(-6)#, about two orders of magnitude smaller than that of #"CO"_3^(2-)#.

Thus, calcium hydroxide dissociates less in water than carbonate associates with water.

That means that the formation of #"Ca"("OH")_2(s)# in water at #25^ @"C"# and #"1 atm"# will be favored, decreasing #["OH"^(-)]# by roughly the ratio of the #K_(sp)# of #"Ca"("OH")_2# and #K_b# of #"CO"_3^(2-)#.

This, in principle, should still yield a #"pH"# higher than #7#, and actually close to around #color(blue)(8)#.

And it apparently yields a #"pH"# of roughly #8.27# with normal #"CO"_2# partial pressures in the air.