# Why does adding water to a saturated solution increase the number of ions present in the solution?

In my book, it is stated that when some water is added to a test tube having a saturated $\ce{Ag2SO4}$ solution, with its solid at equilibrium, the number of $\ce{Ag}$ ions in the solution increases.

I don't understand it. If $\ce{Ag2SO4}$ solution is saturated, it means that no more solute can dissolve, and if we add water, the amount of dissolved solute should not change.

Also, if the amount of dissolved $\ce{Ag}$ increases, would the $K_\mathrm{sp}$ value also increase?

• The concentration is fixed. Not the amount of dissolved ions present in a solution. Commented Mar 8, 2018 at 14:53
• $\pu{K_{sp}}$ is a constant at a given temperature. It doesn’t increase with increase in $\ce{[Ag+]}$ ions. Commented Mar 8, 2018 at 14:57
• Related question here Commented Mar 8, 2018 at 15:07
• The number of $\ce{Ag}$ ions increases, but the number of $\ce{Ag}$ ions per volume of water does not increase. Commented Apr 10, 2019 at 23:23

Consider a saturated solution of $\ce{Ag2SO4}$. The following equilibrium is attained. $$\ce{Ag2SO4 <=> 2Ag+ + SO4^2-}$$
Adding water accounts to increasing the volume of the solution. Hence more amount of solute can be dissolved since solubility of a salt depends on the amount of salt dissolved per unit volume. Therefore, there is an increase in the no. of ions produced in the solution. But as Avnish Kabaj mentioned, $[\ce{Ag+}]$ remains the same.
Also as MollyCooL said, $\pu{K_{sp}}$ doesn't change as it is a constant at a particular temperature.
• @MarvaJami - It depends. Is there more of the solid $\ce{Ag2SO4}$ or not? So if 1 liter dissolves 0.1 grams of something, then 2 liters would dissolve 0.2 grams (total).