The issue of ion concentration in solution mixing

It should be 6g/l and 1g/l, right? I’m not sure if that’s correct

There is an equilibrium relationship in the solution system: Ca2+ + SO42- ⇌ CaSO4. An increase in either the calcium ions or sulfate ions in the solution causes the reaction to proceed in the direction of forming calcium sulfate; as a result, the amount of calcium and sulfate present in the solution in ionic form decreases. Can we still calculate it using the formula (10+2)/2=6?

As the original poster said, sulfate ions react with calcium ions. If we ignore the changes that occur when the volumes are mixed and simply add them together, the solution should originally contain 1 g/L of sulfate ions and 6 g/L of calcium ions. Due to the formation of calcium sulfate, it can be calculated from the reaction equation that all of the sulfate ions are consumed, while the amount of calcium ions consumed is 0.417 g/L; as a result, 5.583 g/L of calcium ions remain in the solution. However, since calcium sulfate is slightly soluble in water, the specific value has to be calculated based on the solubility product

This post was last edited by qugd on 2011-3-31 08:41. It actually deals with the calculation of the remaining sulfate ions after excess sulfate ions form a precipitate with calcium. However, the partial dissolution of calcium sulfate must also be taken into account during the calculation. Due to it being a concentrated salt solution, the salt effect is also quite pronounced. The feasible calculation is carried out in the following steps. My calculation results may not be entirely accurate, but the process should be fine. 1. Apparentally, the mixed solution contains 6 g/L of sulfate ions and 1 g/L of calcium ions ； However, due to the excess of sulfate ions, and since calcium sulfate forms a precipitate, assuming that all calcium ions are precipitated, a total of 3.4 g of calcium sulfate is produced (2.4 g of sulfate ions can be precipitated per gram of calcium). As a result, approximately 3.6 g/L of sulfate ions remain in the solution ； 2. The dissolution of calcium sulfate is taken into consideration; however, due to the common-ion effect, only a very small amount of calcium dissolves in ionic form, which is denoted as X ； Since the solubility product of calcium sulfate is approximately 3.2e-7, and ignoring the effect of salt effects, we have (3.6/96 + X) * X = 3.2e-7. By solving this quadratic equation and selecting the appropriate root, we find that X is approximately equal to 1.7e-5 M ； X represents the calcium ion concentration, approximately 0.0007 g/L ； The solubility of sulfate is slightly over 3.6 g/L. If the salt effect is taken into account, both the sulfate and calcium ion concentrations will increase slightly.