just wondering if someone could tell me that if a large copper sulfate crystal is placed in a hot solution of copper sulfate it is regarded as saturated and also why the solution would darken at first and then become lighter when at equilibrium?
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equilibrium (1 Viewer)
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The solution darkens because you chucked the copper sulfate crystal in, which instantly dissolves and increases the concentration of CuSO4 solution, making the solution a darker blue. I'm not fully sure about the solution getting lighter when at equilibrium, so I'll let someone else answer that. 
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pLuvia
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Is this a practical? or just a general question?lilchezza said:
You increase the solubility of copper sulfate in water by increasing the temperature. So if you heat a saturated solution of copper sulfate and place a large crystal in it more copper sulfate will dissolve in it (like making a super saturated solution) causing it to darken. When it cools down again, the solubility decreases causing some of the copper sulfate to come out of solution forming a solid crystalline structure again causing the sol'n to lighten. The equilibrium exists between the crystal and the saturated solution because the copper sulfate crystal is continually disassociating into Cu<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> ions and likewise these ions are coming out of solution and being deposited back onto the crystal at an proportional rate (CuSO<sub>4</sub>(s) <--> Cu<sup>2+</sup> + SO<sub>4</sub><sup>2-</sup> aka CuSO<sub>4</sub>(aq) )- which is why there will appear to be no change in the appearance of the crystal (actually the blue copper sulfate is hydrate copper sulfate and will be CuSO<sub>4</sub>.5H<sub>2</sub>0 - but for the purposes of this explanation it doesn't make a whole lot of difference).
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