Question

Question #b66ef

Answer 1

Molarity: `5.5 * 10^(-5)"M"`

Explanation:

The trick here is to recognize that you're dealing with an ionic compound that is considered, for all intended purposes, insoluble in aqueous solution.

The molarity of any aqueous solution that contains calcium carbonate will depend on the compound's molar solubility, which you can calculate using its solubility product constant.

The `K_(sp)` for calcium carbonate at `25^@"C"` is listed as being equal to `3.0 * 10^(-9)`.

When you place a solid sample of calcium carbonate in water, you will get a dissociation equilibrium fir which you can use an ICE table to help you determine calcium carbonate's molar solubility.

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

`color(purple)("I")" " " " "" - " " " " " " " " " " ""0" " " " " " " " " " " "0`
`color(purple)("C")" " " "-" " " " " " " " "" (+s)" " " " " " " "(+s)`
`color(purple)("E")" " " "-" " " " " " " " " " ""s" " " " " " " " " " " ""s`

By definition, `K_(sp)` will be equal to

`K_(sp) = s * s = s^2`

This means that the molar solubility of calcium carbonate will be

`sqrt(s^2) = sqrt(3.0 * 10^(-9))`

`s = 5.5 * 10^(-5)"M"`

What this means is that you cannot dissolve more than `5.5 * 10^(-5)` moles of calcium carbonate in one liter of water at `25^@"C"`.

This is equivalent to a mass of

`5.5 * 10^(-5)color(red)(cancel(color(black)("moles"))) * "100.09 g"/(1color(red)(cancel(color(black)("mole")))) = "0.0055 g"`

This means that you cannot dissolve more than 0.0055 g of calcium carbonate in one liter of water.

Youd'd get even less than that in 998 mL

`998color(red)(cancel(color(black)("mL"))) * "0.0055 g"/(1000color(red)(cancel(color(black)("mL")))) = "0.00549 g"`

Most of the solid will remain undissolved. For all intended purposes, the molarity of the solution will be equal to the molar solubility of calcium carbonate

`C = color(green)(5.5 * 10^(-5)"M")`