# The Ksp for BaCO3 is 5.1*10^-9. How many grams of BaCO2 will dissolve in 1000 ml of water?

Nov 22, 2015

$\text{0.014 g}$

#### Explanation:

Te idea here is that you need to use an ICE table to determine barium carbonate's molar solubility, then use the compound's molar mass to determine how many grams will dissolve in that much water.

Barium carbonate, ${\text{BaCO}}_{3}$, is Insoluble in aqueous solution, which means that adding it to water will result in the formation of an equilibrium reaction between the undissolved solid and the dissolved ions.

If you take $s$ to be the molar solubility of calcium carbonate, you can use an ICE table to write

${\text{BaCO"_text(3(s]) " "rightleftharpoons" " "Ba"_text((aq])^(2+) " "+" " "CO}}_{\textrm{3 \left(a q\right]}}^{2 -}$

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

By definition, the solubility product constant, ${K}_{s p}$, will be equal to

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

${K}_{s p} = s \cdot s = {s}^{2}$

This means that the molar solubility of barium carbonate will be equal to

$s = \sqrt{{K}_{s p}} = \sqrt{5.1 \cdot {10}^{- 9}} = 7.14 \cdot {10}^{- 5} \text{M}$

To determine how many grams of barium carbonate you can dissolve in $\text{1.00 L}$ of water, use barium carbonate's molar mass

7.14 * 10^(-5)color(red)(cancel(color(black)("moles")))/"L" * "197.34 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.014 g/L")

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig figs for the volume of the water.