# Is CaCO_3 acidic, basic, or neutral?

Jul 3, 2017

pH = 9.7

#### Explanation:

pH of Saturated $C a C {O}_{3}$ at ${25}^{o} C$:

Jul 3, 2017

It's basic, due to the amount of ${\text{OH}}^{-}$ it generates (greater than ${10}^{- 7}$ $\text{M}$). The $\text{pH}$ should be somewhere near $8$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, with normal ${\text{CO}}_{2}$ partial pressures in the air.

We can examine the dissociation into water using its ${K}_{s p}$:

${\text{CaCO"_3(s) rightleftharpoons "Ca"^(2+)(aq) + "CO}}_{3}^{2 -} \left(a q\right)$

${K}_{s p} = \left[{\text{Ca"^(2+)]["CO}}_{3}^{2 -}\right] = 3.3 \times {10}^{- 9}$

giving each concentration in a saturated solution at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ as

$\left[{\text{Ca"^(2+)] = ["CO}}_{3}^{2 -}\right] = \sqrt{{K}_{s p}}$

$= 5.74 \times {10}^{- 5}$ $\text{M}$

Considering the base's association in water, we obtain the ${K}_{b}$ of ${\text{CO}}_{3}^{2 -}$ from the ${K}_{a}$ of ${\text{HCO}}_{3}^{-}$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$:

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

$= \frac{1 \times {10}^{- 14}}{4.8 \times {10}^{- 11}} = 2.08 \times {10}^{- 4}$

Due to the large ${K}_{b}$, we expect most of the ${\text{CO}}_{3}^{2 -}$ to be gone. The equilibrium association process is

${\text{CO"_3^(2-)(aq) + "H"_2"O"(l) rightleftharpoons "HCO"_3^(-)(aq) + "OH}}^{-} \left(a q\right)$

with the mass action expression

${K}_{b} = \left(\left[{\text{OH"^(-)]["HCO"_3^(-)])/(["CO}}_{3}^{2 -}\right]\right)$

$= {x}^{2} / \left(5.74 \times {10}^{- 5} - x\right) = 2.08 \times {10}^{- 4}$

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 ${\text{CO}}_{3}^{2 -}$ is going to associate.

Solving for the quadratic equation form gives

$2.08 \times {10}^{- 4} \left(5.74 \times {10}^{- 5}\right) - 2.08 \times {10}^{- 4} x - {x}^{2} = 0$

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

$\textcolor{b l u e}{x = 4.68 \times {10}^{- 5}}$ $\textcolor{b l u e}{\text{M}}$,

as the (first) equilibrium concentration of ${\text{OH}}^{-}$, larger than ${10}^{- 7} \text{M}$ (the requirement for basicity at ${25}^{\circ} \text{C}$ and $\text{1 atm}$).

This gives a preliminary $\text{pH}$ of $9.67$, but this is not entirely correct. There are two opposing equilibria for ${\text{OH}}^{-}$ here:

Forward Reaction Forming $\boldsymbol{{\text{OH}}^{-}}$

${\text{CO"_3^(2-)(aq) + "H"_2"O"(l) -> "HCO"_3^(-)(aq) + "OH}}^{-} \left(a q\right)$

Backwards Reaction Forming $\boldsymbol{{\text{OH}}^{-}}$

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

Now, in principle, this ${\text{OH}}^{-}$ would react with the remaining ${\text{Ca}}^{2 +}$ to form "Ca"("OH")_2. That ${K}_{s p}$, however, is actually $5.5 \times {10}^{- 6}$, about two orders of magnitude smaller than that of ${\text{CO}}_{3}^{2 -}$.

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 ${25}^{\circ} \text{C}$ and $\text{1 atm}$ will be favored, decreasing $\left[{\text{OH}}^{-}\right]$ by roughly the ratio of the ${K}_{s p}$ of "Ca"("OH")_2 and ${K}_{b}$ of ${\text{CO}}_{3}^{2 -}$.

This, in principle, should still yield a $\text{pH}$ higher than $7$, and actually close to around $\textcolor{b l u e}{8}$.

And it apparently yields a $\text{pH}$ of roughly $8.27$ with normal ${\text{CO}}_{2}$ partial pressures in the air.