# Question Video: Calculating the Mass of Zinc Carbonate that Will Dissolve in 1000 mL of Water Chemistry

Taking the solubility product of zinc carbonate to be 1.46 × 10⁻¹¹ mol²⋅L⁻² at 298 K, how many grams of zinc carbonate, with molar mass 125.38 g/mol, will dissolve in 1,000 mL of water? Give your answer in scientific notation to 2 decimal places.

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### Video Transcript

Taking the solubility product of zinc carbonate to be 1.46 times 10 to the negative 11 moles squared times liters to the negative two at 298 kelvin, how many grams of zinc carbonate, with molar mass 125.38 grams per mole, will dissolve in 1,000 milliliters of water? Give your answer in scientific notation to two decimal places.

Let’s have a look at the concept of a solubility product before we approach this problem. The solubility product, abbreviated as 𝐾 sp, is the product of the ion concentrations in a saturated solution raised to the power of their respective stoichiometric coefficients. For a generic insoluble compound MA, we can write the dissolution equilibrium with M+ and A− ions as products. Lowercase 𝑚 and 𝑎 represent the stoichiometric coefficients of the ions. This equilibrium reaction has the solubility product equation of 𝐾 sp equals the concentration of M+ times the concentration of A−, with each raised to the power of its respective stoichiometric coefficient.

Let’s look at the relationships between 𝐾 sp and the product of the ion concentrations. When the product of the ion concentrations is equal to the value of the 𝐾 sp, the solution is saturated and the maximum amount of compound that can be dissolved has been reached. When the product of the ion concentrations is greater than the value of the 𝐾 sp, the ions will form a precipitate. When the 𝐾 sp is greater than the product of the ion concentrations, more of the solid can still dissolve.

We are trying to find the maximum amount in grams of the compound that will dissolve until saturated in 1,000 milliliters of water. So, we can use the equation of the solubility product of the compound to first calculate the ion concentrations. Next, using the calculated ion concentrations, we can calculate the number of moles of ions and then the number of moles of solid compound that will dissolve. Finally, using the number of moles of compound and the molar mass of the compound, we can calculate the mass of the compound that will dissolve.

The compound we are given is zinc carbonate, which has the chemical formula ZnCO3. Zinc carbonate dissociates into its constituent ions in the given balanced equilibrium equation. We can use this equation to write the 𝐾 sp for zinc carbonate. This gives us the 𝐾 sp as equal to the concentration of zinc ions times the concentration of carbonate ions. All species in this equilibrium equation have a stoichiometric coefficient of one. This tells us that for every one mole of zinc carbonate dissolved, one mole of zinc ions and one mole of carbonate ions are produced.

Because the zinc and carbonate ions are stoichiometrically equivalent, their concentrations will be equal once the equilibrium is reached. So, we can rewrite the 𝐾 sp as equal to the zinc ion concentration squared. We have been given the value of the 𝐾 sp, so we can substitute this value into the expression. To solve for the zinc ion concentration, we must take the square root of both sides. This gives us the value for the concentration of zinc ions.

Now that we have the ion concentration, we can begin the next step to find the number of moles of dissolved solid. We now know the concentration in moles per liter and we are given the volume, which we can use in this formula where concentration equals the number of moles divided by volume in liters. The volume we are given is in units of milliliters, so we can convert this into liters by dividing by 1,000 since there are 1,000 milliliters in one liter. Milliliters cancel, and this gives us a volume of one liter.

We can rearrange the formula to solve for the number of moles of zinc ions. The formula we will use is the number of moles equals concentration times volume. When we multiply the concentration times the volume, liters will cancel, and we find that the number of moles of dissolved zinc ions is 3.82 times 10 to the negative six. This is the number of moles of zinc ions at equilibrium, but we need to know the number of moles of zinc carbonate that would dissolve. Since these species are stoichiometrically equivalent, we know that for every one mole of zinc ions produced, one mole of zinc carbonate was dissolved. Therefore, the number of moles of zinc carbonate dissolved is the same as the number of moles of zinc ions produced.

In our final step, we will use the formula where molar mass equals mass divided by the number of moles. We now know the number of moles of zinc carbonate dissolved and we are given the molar mass in the question. We can rearrange the formula to solve for mass in grams. The mass will be equal to the number of moles times the molar mass given. Moles will cancel, leaving us in units of grams. We find that the mass is 4.79 times 10 to the negative four. Our answer is already in scientific notation, and we must round to two decimal places.

So, the mass in grams of zinc carbonate that will dissolve in 1,000 milliliters of water is 4.79 times 10 to the negative four grams.