Question

Is `CaCO_3` acidic, basic, or neutral?

Answer 1

pH = 9.7

Explanation:

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

Answer 2

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.

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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.