Here it says determine molar solubility of copper 2 carbonate in .15 molar magnesium carbonate solution. All right, so here they're giving us the KSP of copper 2 carbonate. So that's what we're going to break up into ions Here, Copper to carbonate breaks up into copper 2 ion plus the carbonate ion.
Step one, we set up the ice chart with solid as the only reactant and we cross out the reactant side because remember in an ice chart we ignore solids and liquids. With a nice chart, we have Initial change Equilibrium Step 2 using initial row place the amount given for the common ion. Alright, so let's go back to this question. It says that this is not breaking up in pure water, it's breaking up in a solution that's .15 molar magnesium carbonate.
So magnesium carbonate is made-up of magnesium ion plus the carbonate ion. Each one has a concentration of 0.15 molar. Now where is the common ion? Well, in our ice chart we're dealing with carbonate. Our solution has carbonate. They have the same ion. So that means initially we're starting out with 0.15 molar. We don't have a common ion for copper 2, so it's 0 initially.
Now remember we lose reactants to make product, so we're going to say using the change row, place a + X for the products. So this is plus X + X. Now using the equilibrium row, set up the equilibrium constant expression with KSP and solve for X. The variable X and its number can be ignored if it follows a real number. So let's fill out the rest of this ice chart. Bring down everything. This is plus X. This is .15 + X.
Now it's this portion that we're talking about, because if we look, we have .15 + X. The variable X in its number can be ignored if it follows a real number. This X here follows a real number .15. That means we can ignore this X. And that's because a comparison of point 15X will be so small that it's not going to have an impactful change on the .15 number.
Now here we're going to say see KSP equals products overreactants. Your reactant gets cancelled out because of the solid, so we have copper 2 plus times carbonate here. KSP is 2.4×10-10 = X times 0.15. Again, we're ignoring the plus X. Divide both sides by .15 so we already have our x, x equals 1.6×10-9 molar. This will be our answer because remember when it comes to the solubility or molar solubility of your ionic compound as a whole, that's just X.
So that's what we just found. And step five, we don't need to do it because it says convert found value of X into appropriate units if necessary. Here they want us to find molar solubility which is typically in units of molarity. So we're already done. So this would be our final answer for the molar solubility of our ionic solid.