Is $C a C O_{3}$ $CaCO_3$ acidic, basic, or neutral?

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Is $C a C O_{3}$ $CaCO_3$ acidic, basic, or neutral?

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Answer 1 · 2017-07-03T03:16:49

pH = 9.7

Explanation:

pH of Saturated $C a C O_{3}$ $CaCO_3$ at $25^{o} C$ $25^oC$ :

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Answer 2 · 2017-07-03T08:11:53

It's basic, due to the amount of ${OH}^{-}$ $"OH"^(-)$ it generates (greater than $10^{- 7}$ $10^(-7)$ $M$ $"M"$ ). The $pH$ $"pH"$ should be somewhere near $8$ $8$ at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ , with normal ${CO}_{2}$ $"CO"_2$ partial pressures in the air.

We can examine the dissociation into water using its $K_{s p}$ $K_(sp)$ :

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

$K_{s p} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = 3.3 \times 10^{- 9}$ $K_(sp) = ["Ca"^(2+)]["CO"_3^(2-)] = 3.3 xx 10^(-9)$

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

$[{Ca}^{2 +}] = [{CO}_{3}^{2 -}] = \sqrt{K_{s p}}$ $["Ca"^(2+)] = ["CO"_3^(2-)] = sqrt(K_(sp))$

$= 5.74 \times 10^{- 5}$ $= 5.74 xx 10^(-5)$ $M$ $"M"$

Considering the base's association in water, we obtain the $K_{b}$ $K_b$ of ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ from the $K_{a}$ $K_a$ of ${HCO}_{3}^{-}$ $"HCO"_3^(-)$ at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ :

$K_{b} ({CO}_{3}^{2 -}) = \frac{K_{w}}{K_{a} ({HCO}_{3}^{-})}$ $K_b("CO"_3^(2-)) = K_w/(K_a("HCO"_3^(-)))$

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

Due to the large $K_{b}$ $K_b$ , we expect most of the ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ to be gone. The equilibrium association process is

${CO}_{3}^{2 -} (a q) + H_{2} O (l) ⇌ {HCO}_{3}^{-} (a q) + {OH}^{-} (a q)$ $"CO"_3^(2-)(aq) + "H"_2"O"(l) rightleftharpoons "HCO"_3^(-)(aq) + "OH"^(-)(aq)$

with the mass action expression

$K_{b} = \frac{[{OH}^{-}] [{HCO}_{3}^{-}]}{[{CO}_{3}^{2 -}]}$ $K_b = (["OH"^(-)]["HCO"_3^(-)])/(["CO"_3^(2-)])$

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

Due to the small concentration and the not small $K_{b}$ $K_b$ , we cannot make the small $x$ $x$ approximation on this, as the less of it there is, the more of the ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ is going to associate.

Solving for the quadratic equation form gives

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

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

$x = 4.68 \times 10^{- 5}$ $color(blue)(x = 4.68 xx 10^(-5))$ $M$ $color(blue)("M")$ ,

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

This gives a preliminary $pH$ $"pH"$ of $9.67$ $9.67$ , but this is not entirely correct. There are two opposing equilibria for ${OH}^{-}$ $"OH"^(-)$ 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.](https://en.wikipedia.org/wiki/Calcium_carbonate#With_varying_CO2_pressure)

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