Question

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`

Answer 1

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 `"0.013 g"` of dissociated salt for every `"1 L"` of water at `25^@"C"` `->` see here.

So assuming that your solution is at `25^@"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.

`"CaCO"_ (3(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-)`

This means that the saturated solution of calcium carbonate will contain `3.25 * 10^(-6)` moles of carbonate anions in `"25 mL"` of water, which is essentially equivalent to saying in `"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

`"CO"_ (3(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCO"_ (3(aq))^(-) + "OH"_ ((aq))^(-)`

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

`K_ ("a HCO"_ 3^(-)) = 4.8 * 10^(-11)`

which means that the base dissociation constant of the carbonate anions, `K_ ("b CO"_ 3^(2-))`, will be

`K_ ("b CO"_ 3^(2-)) = (1 * 10^(-14))/(4.8 * 10^(-11)) = 2.08 * 10^(-4)`

This is the case because an aqueous solution at `25^@"C"` has `K_a * k_b = 1 * 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` `"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` `"M"` of hydroxide anions and of bicarbonate anions, it must consume `x` `"M"` of carbonate anions.

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

`K_ ("b CO"_ 3^(2-)) = (x * x)/(1.30 * 10^(-4) - x)`

`2.08 * 10^(-4) = x^2/(1.30 * 10^(-4) -x)`

Rearrange to quadratic equation form

`x^2 + 2.08 * 10^(-4) * x - 2.08 * 1.30 * 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 * 10^(-5)`

This means that, at equilibrium, the resulting solution has

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

Finally, the `"pH"` of the solution will be equal to

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

`"pH"= 14 + log(9.06 * 10^(-5))`

`color(darkgreen)(ul(color(black)("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.