Video: Assessing the Consequences of Adding Sodium Sulfate to the Dissolution of CaSO₄ in Water

Which of the following will not happen when sodium sulfate is added to a saturated solution of CaSO₄ that is at equilibrium? CaSO₄(s) ⇌ Ca²⁺(aq) + SO₄²⁻(aq). [A] The position of equilibrium will shift to the left. [B] The Kₛₚ value will change. [C] The solubility of the calcium sulfate will decrease. [D] The position of equilibrium will shift to resist the increase in sulfate ion concentration. [E] The concentration of calcium ions will decrease.

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Video Transcript

Which of the following will not happen when sodium sulfate is added to a saturated solution of CaSO₄ that is at equilibrium? CaSO₄ solid in equilibrium with Ca²⁺ aqueous plus SO₄²⁻ aqueous. A) The position of equilibrium will shift to the left. B) The K SP value will change. C) The solubility of the calcium sulfate will decrease. D) The position of equilibrium will shift to resist the increase in sulfate ion concentration. Or E) The concentration of calcium ions will decrease.

The question is looking for the one statement that doesn’t happen. So we expect the other four to happen. The symbol for sodium sulfate is Na₂SO₄, and the name for the chemical with symbol CaSO₄ is calcium sulfate. Calcium sulfate is much less water soluble than sodium sulfate is. That extra charge on the calcium ion means that it holds onto the sulfate more strongly than the equivalent sodium ions. So we can imagine our starting scenario is some water with added some calcium sulfate powder. And the calcium sulfate has dissolved, releasing calcium and sulfate ions into solution until the solution is saturated. When a solution is saturated, it means no more of the particular material will dissolve. So in this scenario, what’s going to happen when we add in the much more soluble sodium sulfate?

The sodium sulfate is going to quickly dissolve. The sodium sulfate being more soluble is going to split into sodium plus ions and SO₄²⁻ ions. Now, we’ve looked at the scenario. Let’s have a look at the five statements and see which one does not apply. Statement A suggests that when we add our sodium sulfate, the position of the equilibrium for the dissolution of calcium sulfate will shift to the left. By this, we mean that we’re going to form more calcium sulfate solid and less calcium two plus ions and less sulfate ions from the calcium sulfate. To help us with this one, we can fall back on Le Chatelier’s principle, which tells us that for a dynamic equilibrium, the equilibrium will shift to resist a change in conditions.

For this equilibrium, the only change of any consequence is that, by adding sodium sulfate, we’re increasing the concentration of sulfate ions. Adding more sulfate ions increases the rate at which calcium ions and sulfate ions recombine in solution to form solid calcium sulfate. This shifts the position of equilibrium towards the solid away from the dissolved ions. So statement A is true, but it’s not a correct answer because we’re looking for something that does not happen, not something that does. Statement B says that when we add our solid sodium sulfate to our saturated solution of calcium sulfate, the K SP value will change. K SP stands for the solubility product. And here, it refers to the solubility product of calcium sulfate.

The solubility product is the equilibrium constant for the dissolution of calcium sulfate. So it’s equal to the concentration of calcium ions and the concentration of sulfate ions multiplied together. The K SP value is measured for a salt when it’s on its own. It’s a fundamental property of the salt. So it doesn’t change when we add other materials because that’s not how it’s defined. The clue is in the name, equilibrium constant. Equilibrium constants can change with temperature, but for most other things they’re fixed. So for this particular equilibrium, the K SP value doesn’t change when we add sodium sulfate, so B is our correct answer. But let’s have a look at the others just in case.

Statement C says that the solubility of calcium sulfate will decrease. We’ve already seen this to be true. Solubility is a practical measure of how much of something you can dissolve in a solvent, in this case, calcium sulfate in water. By adding sodium sulfate to our solution of calcium sulfate, we’re increasing the concentration of sulfate ions and decreasing the solubility of the calcium sulfate. This is an example of the common ion effect. By the time you’ve reached equilibrium, we have less calcium sulfate dissolved in solution. So this is in fact true. Therefore, we can move on to statement D. Statement D says that when we add the sodium sulfate, the position of equilibrium for our dissolution will shift to resist the increase in sulfate ion concentration. We’ve already examined this in statement A. So it’s true and we can move on to the last statement.

Statement E says that the concentration of calcium ions will decrease when we add our sodium sulfate. What’s gonna to happen when we add more sodium sulfate is that as the concentration of sulfate ions in solution increases, each calcium ion is going to have a greater chance of bumping into a sulfate ion forming calcium sulfate. This means we’re going to be using up the calcium sulfate in solution, so the concentration of calcium ions in solution will decrease. So statement E is true as well and is therefore not a correct answer. The only statement that will not happen when sodium sulfate is added to a saturated solution of calcium sulfate at equilibrium is that the K SP value of calcium sulfate will change.

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