If I dissolve a Tums tablet ("1000 mg CaCO"_3) in "25 mL" of water, how do I calculate the "pH" of the resultant suspension? When done experimentally, it was between 9 and 10

Jan 8, 2018

Here's what I got.

Explanation:

There are several important things to keep in mind here. For starters, calcium carbonate is considered insoluble in water.

A saturated solution of calcium carbonate will hold about $\text{0.013 g}$ of dissociated salt for every $\text{1 L}$ of water at ${25}^{\circ} \text{C}$ $\to$ see here.

So assuming that your solution is at ${25}^{\circ} \text{C}$, you can say that it will be able to hold

25 color(red)(cancel(color(black)("mL water"))) * "0.013 g CaCO"_3/(10^3color(red)(cancel(color(black)("mL water")))) = "0.000325 g CaCO"_3

of dissociated calcium carbonate. Use the molar mass of calcium carbonate to convert this to moles

0.000325 color(red)(cancel(color(black)("g"))) * "1 mole CaCO"_3/(100.09 color(red)(cancel(color(black)("g")))) = 3.25 * 10^(-6) quad "moles CaCO"_3

Now, the calcium carbonate that will dissociate will do so to produce calcium cations and carbonate anions in $1 : 1$ mole ratios.

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

This means that the saturated solution of calcium carbonate will contain $3.25 \cdot {10}^{- 6}$ moles of carbonate anions in $\text{25 mL}$ of water, which is essentially equivalent to saying in $\text{25 mL}$ of the solution.

The molarity of the carbonate anions will be equal to

["CO"_3^(2-)] = (3.25 * 10^(-6)quad "moles")/(25 * 10^(-3) quad "L") = 1.30 * 10^(-4) quad "M"

A second important thing to keep in mind here is that the carbonate anions will act as a weak base and react with water to form bicarbonate anions and hydroxide anions

${\text{CO"_ (3(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCO"_ (3(aq))^(-) + "OH}}_{\left(a q\right)}^{-}$

The acid dissociation constant for the bicarbonate anions is equal to--see here.

${K}_{{\text{a HCO}}_{3}^{-}} = 4.8 \cdot {10}^{- 11}$

which means that the base dissociation constant of the carbonate anions, ${K}_{{\text{b CO}}_{3}^{2 -}}$, will be

${K}_{{\text{b CO}}_{3}^{2 -}} = \frac{1 \cdot {10}^{- 14}}{4.8 \cdot {10}^{- 11}} = 2.08 \cdot {10}^{- 4}$

This is the case because an aqueous solution at ${25}^{\circ} \text{C}$ has ${K}_{a} \cdot {k}_{b} = 1 \cdot {10}^{- 14}$.

By definition, the base dissociation constant is equal to

K_ ("b CO"_ 3^(2-)) = (["HCO"_3^(-)] * ["OH"^(-)])/(["CO"_3^(2-)])

If you take $x$ $\text{M}$ to be the equilibrium concentration of hydroxide anions, you can say that at equilibrium, the solution will contain

["HCO"_3^(-)] = ["OH"^(-)] = x quad "M"

and

["CO"_3^(2-)] = (1.30 * 10^(-4) - x) quad "M"

This is the case because in order for the reaction to produce $x$ $\text{M}$ of hydroxide anions and of bicarbonate anions, it must consume $x$ $\text{M}$ of carbonate anions.

Plug this into the expression you have for the base dissociation constant to get

${K}_{{\text{b CO}}_{3}^{2 -}} = \frac{x \cdot x}{1.30 \cdot {10}^{- 4} - x}$

$2.08 \cdot {10}^{- 4} = {x}^{2} / \left(1.30 \cdot {10}^{- 4} - x\right)$

${x}^{2} + 2.08 \cdot {10}^{- 4} \cdot x - 2.08 \cdot 1.30 \cdot {10}^{- 8} = 0$

This quadratic equation will produce two solutions, one positive and one negative, but since $x$ is supposed to represent concentration, you can discard the negative one and say that

$x = 9.06 \cdot {10}^{- 5}$

This means that, at equilibrium, the resulting solution has

["OH"^(-)] = 9.06 * 10^(-5) quad "M"

Finally, the $\text{pH}$ of the solution will be equal to

"pH" = 14 - [-log(["OH"^(-)])]

$\text{pH} = 14 + \log \left(9.06 \cdot {10}^{- 5}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 9.96}}}$

I'll leave the answer rounded to two decimal places, the number of sig figs you have for the volume of water.